Infinite series

Hi all,

I've been working my way through the attached question and I'm stuck on the very last part, part iv) (Deduce the sum of the infinite series...)

The answer is (3/5)ln2

I don't really know where to start, but I'm guessing it has something to do with my answers to previous parts of the question :confused:

Any help would be greatly appreciated!

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Reply 1

Jarred

Original post by a nice man

Any help would be greatly appreciated!

In part (b)(ii), you showed that

\dfrac{1}{2r-1}\left( \dfrac{3}{5} \right) ^{2r-1} = I_{r-1}-I_r

.

So first of all, you'll want to use this fact try and rewrite the

\dfrac{1}{2r-1}\left( \dfrac{9}{25} \right) ^{r}

part of the sum in a similar kind of form and then we shall proceed from there.

Are you familiar with the method of differences for a summation?

Edit: Actually, I should have looked at the earlier parts of the question a bit better, they've already sort of guided you through that process. The best way to start off would be to rearrange the equation from (b)(iii)

(edited 10 years ago)

Reply 2

davros

Original post by a nice man

Any help would be greatly appreciated!

If you've done part (iii) correctly, then the final result follows directly if you can show that

I_n \to 0 \text{ as } n \to \infty

Reply 3

a nice man

Original post by davros

If you've done part (iii) correctly, then the final result follows directly if you can show that

I_n \to 0 \text{ as } n \to \infty

I follow you but I don't know how to show that $I_{n} \mapsto 0$ as $n \mapsto \ \infty$

If

$I_{n} = I_{n-1}- \frac{1}{2n-1} \big( \frac{3}{5} \big)^{2n-1}$

How can I show that $I_{n} \mapsto 0$ as $n \mapsto \ \infty$ ?

(edited 10 years ago)

Reply 4

davros

Original post by a nice man

That recurrence relation on its own won't help you.

If you've done part (iii) then you should have that

I_n = ln 2 - (5/3)S_n

where S_n is the series whose limit they want you to find in the final part.

Working back from the answer they've given, if S_n -> (3/5)ln 2 as n->infinity then we must have a limit of 0 for the integral.

So the problem becomes, how can we estimate this integral? We know that tanh x lies between -1 and +1 but this is not quite good enough. Can you think of a better bound on the value of tanh over the range of the integral (it may help to sketch a graph)?

Reply 5

a nice man

Original post by davros

Working back from the answer they've given, if S_n -> (3/5)ln 2 as n->infinity then we must have a limit of 0 for the integral.

So the problem becomes, how can we estimate this integral? We know that tanh x lies between -1 and +1 but this is not quite good enough. Can you think of a better bound on the value of tanh over the range of the integral (it may help to sketch a graph)?

I understand what you mean but in the question the answer isn't given. I just included it so anyone trying to help wouldn't spend ages arriving at the wrong answer.

How could I go about showing $I_{n} \rightarrow 0$ as $n \rightarrow \infty$ without knowing the answer in advance?

Reply 6

davros

Original post by a nice man

Since we have (something)^n inside the integral it helps if we can make "something" less than 1 so that as n gets larger and larger then the integral gets smaller and smaller.

Any bound less than 1 will work, but personally I would say that since tanh is increasing, in the interval (0, ln2) we have tanh x <= tanh(ln2) = 3/5. Therefore we have

\displaystyle 0 \leq \int_0^{ln 2} (tanh \theta)^{2n} d\theta \leq \int_0^{ln 2} (3/5)^{2n} d\theta = [(3/5)^{2n}]ln 2

Therefore as n->infinity the integral tends to 0 and the result follows.

Reply 7

a nice man

Original post by davros

\displaystyle 0 \leq \int_0^{ln 2} (tanh \theta)^{2n} d\theta \leq \int_0^{ln 2} (3/5)^{2n} d\theta = [(3/5)^{2n}]ln 2

Therefore as n->infinity the integral tends to 0 and the result follows.

I don't really understand, but I can sort of grasp what you've done (maybe?). Thanks for taking the time to explain to me

Reply 8

davros

Original post by a nice man

I don't really understand, but I can sort of grasp what you've done (maybe?). Thanks for taking the time to explain to me

No problem!

I don't know where you've found this question, but your screenshot suggests it is an Associated Examining Board question from 1998 which I think puts it just before modular A levels came in, so it's probably a bit tougher than a modern question (although it's possible MEI do things like this - I haven't seen any of their recent textbooks).

Reply 9

a nice man

Original post by davros

The question is from this textbook

http://www.amazon.co.uk/Further-Pure-Mathematics-Brian-Gaulter/dp/0199147353/ref=sr_1_1?ie=UTF8&qid=1391959022&sr=8-1&keywords=further+pure+mathematics

which was published in 2001, explaining the old questions

I think I understand your answer better now. I've attached my working of your method to this post as I can't use LaTex on here to save my life

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Reply 10

davros

Original post by a nice man

Looks good to me :smile:

Some of the older A level maths questions (pre-modular) are pretty scary!!

Reply 11

Jarred

Original post by a nice man

I don't really understand, but I can sort of grasp what you've done (maybe?). Thanks for taking the time to explain to me

It follows from something called the Sandwich Theorem. As far as I know from all the exam boards I've encountered, this isn't something you need to know for the modern A-Level, and is encountered at undergrad these days.

But what it says is that if you three functions satisfying

h(x) \leq f(x) \leq g(x)

and you know that

lim_{x \rightarrow a} h(x) = lim_{x \rightarrow a} g(x) = L

then

lim_{x \rightarrow a} f(x)

is equal to L too.

What I've written above isn't the fully rigorous statement of the theorem, but is enough to give you the general idea of what's going on if you think of it as having two functions that converge to the same point, and then a third function which is bounded between these two functions, intuitively, it must also converge to the same point else it wouldn't be between them.

We basically then set

h=0

f= \int_0^{ln 2} (tanh \theta)^{2n} d\theta

and

g=\int_0^{ln 2} (3/5)^{2n} d\theta = [(3/5)^{2n}]ln 2

and the result follows from applying the theorem.

(edited 10 years ago)

Reply 12

user2020user

Original post by Jarred

...

I was just thinking that this is the "squeeze theorem" I saw a video of ages ago!

Original post by a nice man

...

Another great question nice man!

(edited 10 years ago)

Reply 13

Jarred

Original post by Khallil

I was just thinking that this is the "squeeze theorem" I saw a video of ages ago!

Yeah that's right. I usually call it the Sandwich Theorem but Squeeze Theorem is probably a more common name, it seems to have a million different names depending on where you look,. "Two policemen and a drunk theorem" is probably the most interesting variation :tongue: