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I don’t know mathematics well enough and I was faced with the task of rolling up such a thing. Could you tell me how to do this correctly? $$ \frac{3}{x^3 + 1} + \frac{5}{x^3 + 1} + ... + \frac{2x + 1}{x^3 + 1} $$

Original task

ZiEnTenIn
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    Should the numerator of the last term be $2n+1$ instead of $2x+1$? – whpowell96 Oct 30 '23 at 18:46
  • @whpowell96 no, the unknown is the same in the numerator and denominator – ZiEnTenIn Oct 30 '23 at 18:51
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    Hi, welcome to Math SE. Hint: if $x$ is an integer $\ge1$, $3+5+\cdots2x+1$ is a sum of $x$ terms averaging to $x+2$, so... – J.G. Oct 30 '23 at 18:53
  • Can you clarify whether $x$ is restricted to be an integer? Otherwise it is a bit unclear what the intermediate terms should be – whpowell96 Oct 30 '23 at 18:54
  • @whpowell96 i update question – ZiEnTenIn Oct 30 '23 at 18:59
  • I understood correctly that this is still the same arithmetic progression, which will collapse like $ \frac{ \frac{3}{x^3+1} + \frac{2x+1}{x^3+1}}{2} * x $ ? – ZiEnTenIn Oct 30 '23 at 19:01
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    To complement what @J.G. said, assuming $x$ is a positive integer, put everything over the same common denominator, after which you have a finite arithmetic series in the numerator. – Dave L. Renfro Oct 30 '23 at 19:03
  • @DaveL.Renfro Yes, but in order to calculate the limit I need to first represent the sum by the expression – ZiEnTenIn Oct 30 '23 at 19:07
  • The series converges for $x \rightarrow \infty$ and the result is about $2.909791218786756475847\ldots$ (see https://www.wolframalpha.com/input?i=sum+%28%282x%2B1%29%2F%28x%5E3%2B1%29%29%2C+x%3D1+to+infinity). – Marco Ripà Oct 30 '23 at 19:08
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    I think this will help.... $\sum_\limits{k=1}^n 2k-1 = n^2$ – user317176 Oct 30 '23 at 19:09
  • @MarcoRipà That's not the same problem. – J.G. Oct 30 '23 at 19:16
  • but in order to calculate the limit I need to first represent the sum by the expression --- Google "arithmetic series" + sum. Several of the highest hits, at least for me, seem appropriate, such as this web page (internet archive version, in case the link goes dead in the future sometime). – Dave L. Renfro Oct 30 '23 at 19:19
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    Hint $3+5+\cdots +(2n+1) < (2n+1) + (2n+1)+\cdots (2n+1) = n(2n+1) = 2n^2+2n$ and if you divide quadratic polynomial by a cubic polynomial the limit would be ... – jjagmath Oct 30 '23 at 19:23
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    @jjagmath We agree that the limit is trivially equal to zero, but I do not understand why $n(2n+1)=2n^2+2n$ instead of $n(2n+1)=2n^2+n$ (I assume it is just a typo). – Marco Ripà Oct 30 '23 at 19:26
  • Yes, it's a typo, thank you. – jjagmath Oct 30 '23 at 19:31
  • @MarcoRipà using the hint given to you by user317176, you can see that the numerator is equal to $(x+1)^2$. What you have then is $\frac{(x+1)^2)}{x^3+1} = \frac{x^2+2x+1}{x^3+1}=\frac{x^2}{x^3+1}+\frac{2x}{x^3+1}+\frac{1}{x^3+1}$... can you see why the limit is $0$ now? – H. sapiens rex Oct 30 '23 at 22:11
  • @H.sapiensrex I think you misunderstood the meaning of what I wrote above... I am not going to do basic high school homework here, I simply answered to a different question because the original question wasn't clear enough when you read it in a hurry. My point is that it is trivial to explain why $|\frac{P(x^2)}{P(x^3)}| \rightarrow 0$ as $x \rightarrow \infty$. Regards, M – Marco Ripà Oct 30 '23 at 23:01
  • @MarcoRipà whoops, my apologies! I mistook your comment for one of OP's – H. sapiens rex Oct 30 '23 at 23:27
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    @H.sapiensrex Oh, I see now. No problem at all! – Marco Ripà Oct 30 '23 at 23:39
  • @H.sapiensrex I'm understood, thank you – ZiEnTenIn Oct 31 '23 at 06:14

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