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    Hi, I'm struggling with proving a series converges and finding its limit from the definition of convergence of a series.

    \displaystyle\sum_{i=1}^{\infty} \frac{1}{n(n+2)}

    I understand for the series to converge, the sequence of partial sums must converge so..

    S_N =\displaystyle\sum_{n=1}^{N} \frac{1}{n(n+2)} EDIT: I think this should be different although I'm not sure what it should be

    So now I've converted this into partial fractions and believe I should find the limit as n tends to infinity..

    \displaystyle\lim_{x\to \infty} \frac{1}{2n}-\frac{1}{2(n+2)}=0

    Thus the series is convergent, but the sum is equal to 3/4 which is not 0, I must've gone wrong somewhere? Thanks!
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    (Original post by Substitution)
    Hi, I'm struggling with proving a series converges and finding its limit from the definition of convergence of a series.

    \displaystyle\sum_{i=1}^{\infty} \frac{1}{n(n+2)}

    I understand for the series to converge, the sequence of partial sums must converge so..

    S_N =\displaystyle\sum_{n=1}^{N} \frac{1}{n(n+2)} EDIT: I think this should be different although I'm not sure what it should be

    So now I've converted this into partial fractions and believe I should find the limit as n tends to infinity..

    \displaystyle\lim_{x\to \infty} \frac{1}{2n}-\frac{1}{2(n+2)}=0

    Thus the series is convergent, but the sum is equal to 3/4 which is not 0, I must've gone wrong somewhere? Thanks!
    You've proven that the nth term of the series tends to 0 as n \to \infty, this is far from showing that it convergences.

    For example, we have \lim_{n \to \infty} \frac{1}{n} = 0 but the series \sum_{1}^{\infty} \frac{1}{n} doesn't converge.

    Why don't you just use the comparison test? Show that this series is always smaller than some other convergent series, hint: \sum_{1}^{\infty} \frac{1}{n^2} is convergent.

    Oh and to actually find the limit, write out your partial sums as a telescoping series and use the fact that the nth terms go to infinity as n \to \infty so only the first few terms are relevant (the ones that don't cancel, that is).
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    (Original post by Zacken)
    You've proven that the nth term of the series tends to 0 as n \to \infty, this is far from showing that it convergences.

    For example, we have \lim_{n \to \infty} \frac{1}{n} = 0 but the series \sum_{1}^{\infty} \frac{1}{n} doesn't converge.

    Why don't you just use the comparison test? Show that this series is always smaller than some other convergent series, hint: \sum_{1}^{\infty} \frac{1}{n^2} is convergent.

    Oh and to actually find the limit, write out your partial sums as a telescoping series and use the fact that the nth terms go to infinity as n \to \infty so only the first few terms are relevant (the ones that don't cancel, that is).
    Thanks for the reply. The problem I'm having is not proving the series converges, I think I could prove this with the series convergence tests but the question specifically says to prove the limit exits from the definition and to find it. Would it suffice to state the sequence of partial sums must converge and then write the partial sums as a telescoping series to show the limit exists?

    Thanks
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    By definition the series converges if the partial sums converge. Now to prove that the partial sums converge use a comparison test. Done.
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    (Original post by Substitution)
    Thanks for the reply. The problem I'm having is not proving the series converges, I think I could prove this with the series convergence tests but the question specifically says to prove the limit exits from the definition and to find it. Would it suffice to state the sequence of partial sums must converge and then write the partial sums as a telescoping series to show the limit exists?

    Thanks
    No, don't state the sequence of partials sums converges, prove that it converges using the comparison test and hence the series must comverge, from which telescoping the partials gets you the limit.
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    (Original post by Zacken)
    No, don't state the sequence of partials sums converges, prove that it converges using the comparison test and hence the series must comverge, from which telescoping the partials gets you the limit.
    (Original post by Math12345)
    By definition the series converges if the partial sums converge. Now to prove that the partial sums converge use a comparison test. Done.
    Perfect, Thank you.
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    (Original post by Substitution)
    Perfect, Thank you.
    No problem. :cool:
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    (Original post by Zacken)
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    (Original post by Math12345)
    .
    Sorry to be a pain, can you please check my proof of convergence suffices? or is there something I should add? Thanks!

    EDIT: I just feel like I'm not actually mentioning the partial sums at all.
    (Excuse the awful hand writing)
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    (Original post by Substitution)
    Sorry to be a pain, can you please check my proof of convergence suffices? or is there something I should add? Thanks!

    EDIT: I just feel like I'm not actually mentioning the partial sums at all.
    (Excuse the awful hand writing)
    I'd say that's okay (in my very limited experience), although you want to say that your "sequence of partial sums..." and not "series of partial sums". You also want to say "By the comparison test..." and not "convergence test".

    I would end the final paragraph with something like "since the sequence of partials sums converges to a finite limit as shown by the comparison test, then the series converges".
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    (Original post by Zacken)
    I'd say that's okay (in my very limited experience), although you want to say that your "sequence of partial sums..." and not "series of partial sums". You also want to say "By the comparison test..." and not "convergence test".

    I would end the final paragraph with something like "since the sequence of partials sums converges to a finite limit as shown by the comparison test, then the series converges".
    Thank you so much!!! Appreciate it massively!
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    (Original post by Substitution)
    Thank you so much!!! Appreciate it massively!
    No problem at all. This is good practice for me next year, anyhow. :lol:
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    (Original post by Zacken)
    No problem at all. This is good practice for me next year, anyhow. :lol:
    Are you an A-Level student if you don't mind me asking? You're crazy smart, I'm jealous haha.
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    (Original post by Substitution)
    Are you an A-Level student if you don't mind me asking? You're crazy smart, I'm jealous haha.
    Yeah, I am. I have a feeling I'm going to get rekt by Analysis next year. :lol: How are you finding it so far?
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    (Original post by Substitution)
    Hi, I'm struggling with proving a series converges and finding its limit from the definition of convergence of a series.

    \displaystyle\sum_{i=1}^{\infty} \frac{1}{n(n+2)}

    I understand for the series to converge, the sequence of partial sums must converge so..

    S_N =\displaystyle\sum_{n=1}^{N} \frac{1}{n(n+2)} EDIT: I think this should be different although I'm not sure what it should be

    So now I've converted this into partial fractions and believe I should find the limit as n tends to infinity..

    \displaystyle\lim_{x\to \infty} \frac{1}{2n}-\frac{1}{2(n+2)}=0

    Thus the series is convergent, but the sum is equal to 3/4 which is not 0, I must've gone wrong somewhere? Thanks!
    Do you know what converge means? From what you've written, you are not making any sense.
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    (Original post by Zacken)
    Yeah, I am. I have a feeling I'm going to get rekt by Analysis next year. :lol: How are you finding it so far?
    I'm sure you'll be fine! it's nowhere near as hard as people make out, personally I find these series the hardest topic studied at first year. Where are you hoping to go to uni?
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    (Original post by Substitution)
    I'm sure you'll be fine! it's nowhere near as hard as people make out, personally I find these series the hardest topic studied at first year. Where are you hoping to go to uni?
    Hardest topic in analysis or in general from the first year? If I meet my offer (hopefully) Cambridge, otherwise it's Warwick. What uni are you at? (if you don't mind me asking)
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    (Original post by Pablo Picasso)
    Do you know what converge means? From what you've written, you are not making any sense.
    If you look above, I posted a picture of the proof after a bit of help from Zacken. If you wouldn't mind double checking that, it would be appreciated. I don't need to incorporate the sequence of partial sums more than I have do I?

    (Original post by Zacken)
    Hardest topic in analysis or in general from the first year? If I meet my offer (hopefully) Cambridge, otherwise it's Warwick. What uni are you at? (if you don't mind me asking)
    Honestly, I would say in general. With that being said, it's the hardest topic although I find proofs of theorems much harder but these are incorporated into tests much less. Sounds good, best of luck with your A-Levels - I'm at Birmingham
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    (Original post by Substitution)
    Honestly, I would say in general. With that being said, it's the hardest topic although I find proofs of theorems much harder but these are incorporated into tests much less. Sounds good, best of luck with your A-Levels - I'm at Birmingham
    I would have thought the sort of stuff like group theory and algebra would have been the hardest (in my opinion, but that's maybe because I suck really badly at geometry, so...). Thanks, good luck with your uni exams as well.
 
 
 
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