Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    2
    ReputationRep:
    (Original post by joostan)
    OK, I won't.
    Is that problem 161? Looks like a mess, but I might take a punt, right now I'm having a go at some of the others that you just posted
    It's actually really neat and, dare I say, 'straightforward' provided that you have the right tools. The hardest part is having the determination to persist (because everything seems to feel like a dead end in this problem - even if it's along the right lines!)
    To demonstrate that these are the only solutions with the given restrictions..
    Nice solution bro, but why not simply use an irrationality argument? The question is far more straightforward that its 'BMO2'-status suggests!
    (Original post by FireGarden)
    1^{\infty} = \displaystyle\lim_{x\to\infty} \left(1+\dfrac{1}{x} \right)^{x} = e
    It just feels that, representing 1^{\infty} in that way assumes it's indeterminate in the first place as you are adding in the "1+\frac{1}{x}" which approaches 1 at a different rate to the constant function "1" itself, hence creating a circular argument.
    (Original post by Smaug123)
    Here's my "progress post", updated as I solve the question

    k=1 case: I used Mathematica to get a couple of hundred iterations, I just couldn't intuit the limit… then it was easy enough to show by induction that b_n \equiv \dfrac{a_n^2}{n} > 2 always, after getting a recurrence relation for b_n. Also shown previously that b_n is strictly decreasing. Now I just need to show that 2 is indeed the limit.
    I wouldn't get too caught up in pedantry! My solution came out of what I initially considered to be a conjecture until I rigorously formalized my proof.
    (Original post by MW24595)
    Solution 307

    Solution 310

      \displaystyle \frac{1}{2} \displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \frac{m^2 n(3^n m)+ n^2 m (3^m n)}{3^{m+n} (n 3^m + m3^n)} \right)
    Awesome solutions bro, I did the same in both cases. Also, this line in a typo^
    Offline

    11
    ReputationRep:
    (Original post by Jkn)
    It's actually really neat and, dare I say, 'straightforward' provided that you have the right tools. The hardest part is having the determination to persist (because everything seems to feel like a dead end in this problem - even if it's along the right lines!)

    Nice solution bro, but why not simply use an irrationality argument? The question is far more straightforward that its 'BMO2'-status suggests!
    OK, I'll give it a go tomorrow.
    I figured irrationality looked like a faff of writing words, to be honest the quadratic looks messy but is actually quite simple
    I've spent a while on Problem 305.
    Spoiler:
    Show
    I've figured out that it's every other Fibonacci number, and by using the Fibonacci sequence I can demonstrate that each term is an integer, but I think I've made a slip somewhere, and can't seem to spot it
    I'll give it a go again a bit later
    Offline

    2
    ReputationRep:
    (Original post by joostan)
    OK, I'll give it a go tomorrow.
    I figured irrationality looked like a faff of writing words, to be honest the quadratic looks messy but is actually quite simple
    I've spent a while on Problem 305.
    Spoiler:
    Show
    I've figured out that it's every other Fibonacci number, and by using the Fibonacci sequence I can demonstrate that each term is an integer, but I think I've made a slip somewhere, and can't seem to spot it
    I'll give it a go again a bit later
    Spoiler:
    Show
    Yeah that's right! All that is required is a proof by induction that the sub-sequence of the Fibonacci numbers satisfy the recurrence relation.

    This can be spotted very quickly by considering what's special about \sqrt{5x^4 \pm 4}.
    Offline

    0
    ReputationRep:
    (Original post by Jkn)

    Problem 308
    *

    Find all pairs of integers (x,y) such that \displaystyle 1+x^2y=x^2+2xy+2x+y

    Problem 314*

    Find all positive integers n such that n+2012 divides n^2+2012 and n+2013 divides n^2+2013.
    Solution 308

    The first thing one should note, is that, considering the equation  \ mod \ x , we have,  y \equiv 1 \ mod \ x \Rightarrow y-1 \equiv 0 \ mod \ x. This is, ofcourse, considering that  x \not= 0 .

    If it is, then simple substitution gives us, y= 1. So,  (0,1) is one solution. Let us now assume that x is non-zero.

    Now,

     1+x^2 y = x^2+ 2 xy+ 2x + y \Rightarrow (x^2-1)(y-1) = 2x (y+1) \Rightarrow (x-1)(x+1)(y-1)= 2x (y+1) .

    As,  x | y-1 , let  y-1 = kx for some,  k \in \mathbb{Z} .

     \Rightarrow kx(x-1)(x+1) = 2x (kx + 2) \Rightarrow k(x-1)(x+1) = 2(kx+2)



\Rightarrow kx^2 - 2kx - k - 4 = 0



\Rightarrow x=\displaystyle \frac{2 \pm \sqrt{4k^2 + 4k(k+4)}}{2k} = 1 \pm \sqrt{1+ \frac{k+4}{k}}



\Rightarrow k| k+4 \Rightarrow k|4 \Rightarrow k = \pm 1, \pm 2, \pm, 4

    Substituting these into the expression for  x, we see that only  k = -4, 2, 2 yield integer values for x .

    So, we have possible pairs,  (k,x) = (-4, 0), (-4, 2), (-2, 1), (2, -1), (2, 3) .

    Computing  y from these, we have,  (x, y)= (0,1), (2, -7), (-1, 1), (3, 7) , and, these are the only possible pairs of integers satisfying the equation above.

    Solution 314

    I'm posting this, because I think mine's one way of looking at it, but I'm concealing it because it's a nice one that I hope others solve as well.

    Spoiler:
    Show

    Note, that all integers here are positive.
    Let n^2 + 2012 = k(n+2012), n^2+ 2013 = m(n+2013) . Then, subtracting the 2, we have,

     \Rightarrow 1 = m(n+2013) - k(n+2012) = (m-k)(n+2012) + m

    From the second equation we have,  m < n+ 2013 \Rightarrow m =k = 1.

     \Rightarrow n = 1 is the only solution.

    .
    Offline

    0
    ReputationRep:
    (Original post by Jkn)
    Problem 305*

    Prove that the sequence defined by y_0=1, \displaystyle y_{n+1}=\frac{1}{2} \left(3y_n + \sqrt{5y_n^2-4} \right) for n \ge 0 consists only of integers.

    [As a bonus, can you find any 'similar' sequences for which this is true?]
    I'll wait for others to post a solution to this one first, but I just wanted to point out something.

    Did anybody else notice that  y_n = F_{2n+1} , where  F_n denotes the nth Fibonacci number?
    So, this sequence, essentially generates every other Fibonacci number.

    Remember that as  n \to \infty, \displaystyle F_{n+1} \cong \frac{1+ \sqrt {5}}{2} F_n .

    Now, futher note that, as,  n \to \infty, \displaystyle y_{n+1} \cong \frac{3+ \sqrt{5}}{2} y_n .

    And finally, notice that,  \displaystyle \frac{3+ \sqrt{5}}{2} = \left(\frac{1+ \sqrt {5}}{2} \right)^2 .

    (Original post by Jkn)
    Awesome solutions bro, I did the same in both cases. Also, this line in a typo^
    Cheers, man. These were a good bunch, eh?
    Offline

    0
    ReputationRep:
    (Original post by Jkn)
    Problem 312*

    Evaluate \displaystyle \frac{4}{4^2-1}+\frac{4^2}{4^4-1}+\frac{4^4}{4^8-1}+\frac{4^8}{4^{16}-1}+\cdots
    An ugly solution, but a solution nonetheless.

    Let  \displaystyle S_n(x) = \frac{x}{x^2-1}+\frac{x^2}{x^4-1}+\frac{x^4}{x^8-1}+\frac{x^8}{x^{16}-1} + ... + \frac{ x^{2^n} }{ x^{2^{n+1}} -1}

    After adding a couple of terms I guessed that (and proved it by induction)

    \displaystyle S_n(x) = \frac{x+x^2+x^3+...+x^{2^{n+1}-1}}{x^{2^{n+1}}-1}

    which can be written as \displaystyle S_n(x) = \frac{x}{x-1} \cdot \frac{\frac{1}{x}-\frac{1}{x^{2^{n+1}}}}{1-\frac{1}{x^{2^{n+1}}}} by summing the GP and tidying up a little.

    Taking the limit, we get  \displaystyle \lim_{n \rightarrow \infty} S_n(x)=\frac{1}{x-1}

    So for this question setting x=4 gives the sum as \frac{1}{3}.
    Offline

    0
    ReputationRep:
    (Original post by Jkn)
    Problem 312*

    Evaluate \displaystyle \frac{4}{4^2-1}+\frac{4^2}{4^4-1}+\frac{4^4}{4^8-1}+\frac{4^8}{4^{16}-1}+\cdots

    Solution 312


     \displaystyle \frac{4}{4^2-1}+ \frac{4^2}{4^4-1}+ \frac{4^4}{4^8-1}+\frac{4^8}{4^{16}-1}+ \cdots= \frac{1}{4} \left(1+ \frac{1}{4^2}+ \frac{1}{4^4}+...  \right)+  \frac{1}{4^2} \left(1+ \frac{1}{4^4}+ \frac{1}{4^8}+... \right)+\frac{1}{4^4} \left(1+ \frac{1}{4^8}+ \frac{1}{4^{16}}+... \right)+...
     \displaystyle = \left(\frac{1}{4}+ \frac{1}{4^3}+ \frac{1}{4^5}+... \right)+  \left(\frac{1}{4^2}+ \frac{1}{4^6}+ \frac{1}{4^{10}}+... \right)+ \left(\frac{1}{4^4}+ \frac{1}{4^{12}}+ \frac{1}{4^{20}}+... \right)+...

    Note that all (negative) exponents of 4 in the first bracket are \equiv 1 \mod 2 (moreover, they contain all positive integers of this form), the next bracket has terms \equiv 2 \mod 4, then the third bracket \equiv 4 \mod 8 ...

    Lemma: All positive integers are of the form \equiv 2^{n-1} \mod 2^{n} for some unique n, (proof of n's existence and uniqueness taken to be obviously true, can clarify if unclear).

    As there must exist exactly a single bracket containing the negative exponents of the form \equiv 2^{n-1} \mod 2^{n}, and there exist brackets for each n, all negative exponents but zero exist in the expansion.

    Thus the expansion is equal to
    1+\frac{1}{4}+\frac{1}{4^2}+ \frac{1}{4^3} +...= \frac{1}{1-\frac{1}{4}}-1=\frac{1}{3}
    Offline

    2
    ReputationRep:
    (Original post by MW24595)
    Solution 308
    Looks a tad longer than I had expected. Don't think that was quite how I did it, but good none the less!
    Solution 314

    I'm posting this, because I think mine's one way of looking at it, but I'm concealing it because it's a nice one that I hope others solve as well.
    Again, not how I did it though just as neat. I think there's no surprise you went for divisibility/modular-arithmetic-style solutions whilst I went for good old fashioned algebra in both cases (I should really start trying to produce some solutions more like yours as there may well be cases where your style proves more generalisable).
    (Original post by MW24595)
    Spoiler:
    Show
    Did anybody else notice that  y_n = F_{2n+1} , where  F_n denotes the nth Fibonacci number?


    Cheers, man. These were a good bunch, eh?
    Nice use of spoiler brackets there! :lol: I'm pretty sure that's the key to the problem? There are also several methods that could have been used to spot that. How did you spot it?

    Yeah they are! Some I found rather boring (mostly the BMO1/BMO2 ones as they are often rather trivial and rely on the application of a single technique - one that cannot be known without the specialist knowledge - which kind of frustrates me that I never had the chance to do one of these competitions since I have learnt the techniques :lol:) The one above is fun though, as are all the non-BMO questions (in my opinion)!

    Oh and, btw, I eagerly await your solution to the \displaystyle a_{n+1}=a_{n}+\frac{1}{\sqrt[k]{a_{n}}} question...
    (Original post by bogstandardname)
    An ugly solution, but a solution nonetheless.
    (Original post by henpen)
    Solution 312
    Nice stuff guys, it's great to see a lot of alternative solutions!

    For me, as a lover of direct algebraic solutions, this method simply jumped out at me.

    Solution 312 (3)

    Factorising the denominator, using the method of differences and then noting that the nth term tends to zero, we have:

    \displaystyle \begin{aligned} \sum_{n=0}^{\infty} \frac{(4^{2^{n}}+1)-1}{(4^{2^{n}}+1)(4^{2^{n}}-1)} = \sum_{n=0}^{\infty} \left( \frac{1}{4^{2^{n}}-1} - \frac{1}{4^{2^{n+1}}-1} \right) = \frac{1}{4^{2^{0}}-1}=\frac{1}{3} \ \square \end{aligned}
    Offline

    13
    ReputationRep:
    (Original post by james22)
    Problem 304*

    Does there exist an integer, k, such that using only at most k coin flips, a number in the set {1,2,3} can be randomly picked with each being equally likely?
    (Original post by Jkn)
    Solution 304

    There exists no such integer k....
    If the coin is thick enough to flip heads, tails and side with equal probability then clearly k = 1. Thanks to some physicists from Harvard, one knows exactly how to construct such a coin.
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by ukdragon37)
    If the coin is thick enough to flip heads, tails and side with equal probability then clearly k = 1. Thanks to some physicists from Harvard, one knows exactly how to construct such a coin.
    PRSOM - this is an excellent solution!
    Offline

    11
    ReputationRep:
    (Original post by MW24595)
    I'll wait for others to post a solution to this one first, but I just wanted to point out something.

    Did anybody else notice that  y_n = F_{2n+1} , where  F_n denotes the nth Fibonacci number?
    So, this sequence, essentially generates every other Fibonacci number.
    Yep. I did


    (Original post by Jkn)
    Problem 305*

    Prove that the sequence defined by y_0=1, \displaystyle y_{n+1}=\frac{1}{2} \left(3y_n + \sqrt{5y_n^2-4} \right) for n \ge 0 consists only of integers.

    [As a bonus, can you find any 'similar' sequences for which this is true?]
    Solution 305:
    Lemma: (I hope that's used correctly )
    F_n^2-F_{n+1}F_{n-1}=(-1)^{n-1}
    Where F_n denotes the n^{th} Fibonacci number.
    This is Cassini's Identity.
    Proof:
    Basis Case: F_1^2-F_2F_0=1=(-1)^0
    Now assume that the identity holds for:
    n=k \ k \in \mathbb{N} So that F_k^2-F_{k+1}F_{k-1}=(-1)^{k-1}
    Now consider;
    F_{k+1}^2-F_kF_{k+2}=F_{k+1}^2-F_k(F_{k+1}+F_{k})=F_{k+1}^2-F_kF_{k+1}-F_k^2

\Rightarrow F_{k+1}^2-F_kF_{k+2}=F_{k+1}(F_k+F_{k-1})-F_kF_{k+1}-F_k^2=F_{k+1}F_{k-1}-F_k^2
    Recall that: F_k^2-F_{k+1}F_{k-1}=(-1)^{k-1}
    \Rightarrow F_{k+1}^2-F_kF_{k+2}=(-1)(-1)^{k-1}=(-1)^k
    This completes the induction.

    Now to the problem:
    y_{n+1}=\dfrac{1}{2} \left(3y_n + \sqrt{5y_n^2-4} \right)
    y_0=1 (=F_1)

y_1=2 \ (=F_3)

y_2=5 \ (=F_5)

y_3=13 \ (=F_7)
    This is looking a lot like alternate Fibonacci numbers.
    So the proposition is that: y_n=F_{2n+1}
    We have the Basis Case(s) for n=\{0,1,2,3\}
    So assume that this is true for n=k, \ k \in \mathbb{N}
    So that: y_k=F_{2k+1}
    Now:
    y_{k+1}=\dfrac{1}{2} \left(3F_{2k+1} + \sqrt{5F_{2k+1}^2-4} \right)
    Consider:
    \sqrt{5F_{2k+1}^2-4}
    Observing that by the lemma:
    -4=(-4)(-1)^{2k}=-4(F_{2k+1}^2-F_{2k}F_{2k+2})

\Rightarrow \sqrt{5F_{2k+1}^2-4}=\sqrt{5F_{2k+1}^2-4(F_{2k+1}^2-F_{2k}F_{2k+2})}

\Rightarrow \sqrt{5F_{2k+1}^2-4}=\sqrt{F_{2k+1}^2+4F_{2k}(F_{2  k+1}+F_{2k})}

\Rightarrow \sqrt{5F_{2k+1}^2-4}=\sqrt{F_{2k+1}^2+4F_{2k}F_{2k  +1}+4F_{2k}^2}

\Rightarrow \sqrt{5F_{2k+1}^2-4}=\sqrt{(F_{2k+1}+2F_{2k})^2}=(  F_{2k+1}+2F_{2k})
    Now substituting this in to the recurrence relation yields:
    y_{k+1}=\dfrac{1}{2} \left(3F_{2k+1}+F_{2k+1}+2F_{2k} \right)=2F_{2k+1}+F_{2k}
    \Rightarrow y_{k+1}=(F_{2k+1}+F_{2k})+F_{2k+  1}=F_{2k+2}+F_{2k+1}

\therefore y_{k+1}=F_{2k+3}
    This implies that all y_n=F_{2n+1} \ \forall n \in \mathbb{Z^+}
    Since every term in the Fibonacci sequence is an integer (since an integer added to another integer is itself an integer) it follows that every term in the sequence:
    y_{n+1}=\dfrac{1}{2} \left(3y_n + \sqrt{5y_n^2-4} \right)
    Is also an integer.

    Bonus:
    Spoiler:
    Show
    It can be demonstrated in a similar fashion that the sequence with: y_0=1 defined by the recurrence relation:
    y_{n+1}=\dfrac{1}{2} \left(3F_{2k+1} + \sqrt{5F_{2k+1}^2+4} \right)=F_{2n}
    From this it follows that the test for a Fibonacci number is to see if:
    (5x^2 \pm 4)=n^2 For some n \in \mathbb{Z}
    Offline

    15
    ReputationRep:
    (Original post by ukdragon37)
    If the coin is thick enough to flip heads, tails and side with equal probability then clearly k = 1. Thanks to some physicists from Harvard, one knows exactly how to construct such a coin.
    Nice, but by coin I really meant random binary generator.
    Offline

    0
    ReputationRep:
    Problem 316*

    Prove that
    \displaystyle \sum_{j<k, j,k \in \mathbb{N}}\frac{1}{j^2k^2}= \frac{\pi^4}{120}
    and find a general form for
    \displaystyle \sum_{x_1<x_2<...<x_n, x_1,x_2,...,x_n \in \mathbb{N}}\frac{1}{\prod_{k=1}^  nx_k^2}.

    Note that I'm taking the set of natural numbers to not include 0.

    Spoiler:
    Show
    Consider the roots of the Taylor series for \frac{\sin(\sqrt{y})}{\sqrt{y}}.
    Offline

    0
    ReputationRep:
    (Original post by henpen)
    Problem 316*

    Prove that
    \displaystyle \sum_{j<k, j,k \in \mathbb{N}}\frac{1}{j^2k^2}= \frac{\pi^4}{120}
    and find a general form for
    \displaystyle \sum_{x_1<x_2<...<x_n, x_1,x_2,...,x_n \in \mathbb{N}}\frac{1}{\prod_{k=1}^  nx_k^2}
    Solution 316

    Notice that,

     \displaystyle \sum_{j<k, j,k \in \mathbb{N}}\frac{1}{j^2k^2} = \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2} .

    Now, note that:

     \displaystyle \frac{\sin {\sqrt {y}}}{\sqrt{y}} = \displaystyle \sum_{r=1}^{\infty} (-1)^r \frac{y^r}{2r+1!} .

    Consider the Weierstrauss Product of this function, i.e, when a function is expressed as a product of linear factors, each of which has a root equal to the root of this function.

    Now,   \displaystyle \frac{\sin {\sqrt {y}}}{\sqrt{y}} has roots at  \pm \pi, \pm 2 \pi, \pm 3 \pi ... .

     \Rightarrow \displaystyle \frac{\sin {\sqrt {y}}}{\sqrt{y}} = (1- \frac{\sqrt y}{\pi})(1+ \frac{\sqrt y}{\pi})(1- \frac{\sqrt y}{2 \pi})...



\Rightarrow \displaystyle \frac{\sin {\sqrt {y}}}{\sqrt{y}} = \displaystyle \prod_{i=1}^{\infty} \left(1- \frac{y}{i^2 \pi ^2} \right)



.

    And herein lies the crux move. Consider the co-efficient of  y^2 in this expansion.

    We have:

    Co-efficient of  y^2 = \displaystyle \frac{1}{\pi^4} \left( \left(\frac{1}{2^2}+ \frac{1}{3^2}... \right)+ \frac{1}{2^2} \left(\frac{1}{3^2}+ \frac{1}{4^2}...\right) ... \right) = \frac{1}{\pi^4} \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2}  .

    And voila, this is (almost) the very expression we had in the first line.

    Notice that in the Maclaurin expansion of  \displaystyle \frac{\sin {\sqrt {y}}}{\sqrt{y}} , we had the co-efficient of  y^2 = \frac{1}{5!} .

    Equating the 2, we have:

     \displaystyle\frac{1}{\pi^4} \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2} = \frac{1}{5!}



\Rightarrow \displaystyle \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2} = \frac{\pi^4}{5!}



\Rightarrow \displaystyle \sum_{j<k, j,k \in \mathbb{N}}\frac{1}{j^2k^2} = \frac{\pi^4}{5!}

    A generalization can easily be undertaken considering higher powers of  y in both expressions.

    So, we have,

     \displaystyle \sum_{x_1<x_2<...<x_n, x_1,x_2,...,x_n \in \mathbb{N}}\frac{1}{\prod_{k=1}^  nx_k^2} = \displaystyle \frac{\pi^{2n}}{2n+1!}
    Offline

    0
    ReputationRep:
    (Original post by MW24595)
    Solution 316
    ...
    And herein lies the crux move. Consider the co-efficient of  y^2 in this expansion.

    We have:

    Co-efficient of  y^2 = \displaystyle \frac{1}{\pi^4} \left( \left(\frac{1}{2^2}+ \frac{1}{3^2}... \right)+ \frac{1}{2^2} \left(\frac{1}{3^2}+ \frac{1}{4^2}...\right) ... \right) = \frac{1}{\pi^4} \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2}  .

    And voila, this is (almost) the very expression we had in the first line.

    Notice that in the Maclaurin expansion of  \displaystyle \frac{\sin {\sqrt {y}}}{\sqrt{y}} , we had the co-efficient of  y^2 = \frac{1}{5!} .

    Equating the 2, we have:

     \displaystyle\frac{1}{\pi^4} \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2} = \frac{1}{5!}



\Rightarrow \displaystyle \sum_{i=1}^{\infty} \frac{1}{i^2} \sum_{j=i+1}^{\infty} \frac{1}{j^2} = \frac{\pi^4}{5!}



\Rightarrow \displaystyle \sum_{j<k, j,k \in \mathbb{N}}\frac{1}{j^2k^2} = \frac{\pi^4}{5!}

    ...
    Nice. Consider this way as an alternative to your expansion:

    \displaystyle \frac{\sin(\sqrt{y})}{\sqrt{y}}=  \sum_{k=0}^\infty a_k y^k

    Letting r_k be the roots of the function, we have
    \displaystyle a_0=(-1)^n\prod_{k=0}^\infty r_k=1
    \displaystyle a_1=(-1)^{n-1}\sum_{k=1}^\infty \frac{1}{r_k}\prod_{k=0}^\infty r_k=\frac{-1}{3!}
    \displaystyle a_1=(-1)^{n-1}\sum_{j,k}^\infty \frac{1}{r_kr_j}\prod_{k=0}^ {\infty}   r_k=\frac{1}{5!}
    , where n is an arbitrary integer (basically a_k and a_{k-l} have opposite parity only if l is odd).
    Then take the quotient \frac{a_m}{a_0} and substitute r_k to get the required result.
    Offline

    0
    ReputationRep:
    Problem 317**


    Find

    \displaystyle \sum_{k=-\infty}^{\infty}\frac{1}{2k+1}-\frac{2}{\pi(2k+1)^2}
    Offline

    0
    ReputationRep:
    (Original post by henpen)
    Problem 317**


    Find

    \displaystyle \sum_{k=-\infty}^{\infty}\frac{1}{2k+1}-\frac{2}{\pi(2k+1)^2}
    I'm not quite sure this is right, but...

    Solution 317 (I think?)

    Spoiler:
    Show


     \displaystyle \sum_{k= -\infty}^{\infty} \frac{1}{2k+1} = \displaystyle \sum_{k= -\infty}^{-1} \frac{1}{2k+1} + \displaystyle \sum_{k=0}^{\infty} \frac{1}{2k+1}= ... -\frac{1}{5} - \frac{1}{3} - 1 + 1 + \frac{1}{3} + \frac{1}{5}... = 0



\displaystyle \sum_{k= 1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}



\Rightarrow \displaystyle \sum_{k= 1}^{\infty} \frac{1}{(2k)^2} = \frac{\pi^2}{24}



\Rightarrow \displaystyle \sum_{k= 1}^{\infty} \frac{1}{k^2}- \frac{1}{(2k)^2} =  \displaystyle \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2}= \frac{\pi^2}{8}



\Rightarrow \displaystyle \displaystyle \sum_{k=1}^{\infty} \frac{1}{(2k-1)^2} = \displaystyle \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}= \frac{\pi^2}{8}



\Rightarrow \displaystyle \sum_{k=0}^{\infty} \frac{2}{\pi(2k+1)^2} = \frac{\pi}{4}

    Notice that,

     \displaystyle \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = \displaystyle \sum_{k=-\infty}^{-1} \frac{1}{(2k+1)^2}



\Rightarrow \displaystyle \sum_{k=-\infty}^{\infty}\frac{1}{2k+1}-\frac{2}{\pi(2k+1)^2} = -\frac{\pi}{2}

    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by MW24595)
    Solution 317 (I think?)
    Does this not require absolute convergence? You've split up the summand, which you can't necessarily do without absolute convergence.
    Offline

    0
    ReputationRep:
    (Original post by Smaug123)
    Does this not require absolute convergence? You've split up the summand, which you can't necessarily do without absolute convergence.
    Hence my hesitation. In fact, I don't think it is.(?)
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by MW24595)
    Hence my hesitation. In fact, I don't think it is.(?)
    I'm not sure - it feels not-absolutely-convergent, but I don't know. Mathematica can't make heads or tails of the absolute series - refuses to even give me a numerical approximation.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Has a teacher ever helped you cheat?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Write a reply...
    Reply
    Hide
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.