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This question already has an answer here:

Could someone please help me with how do I calculate the sum of the $$\sum_{n=1}^{\infty}\frac{1}{4n^{2}-1}$$ infinite series? I see that $$\lim_{n\rightarrow\infty}\frac{1}{4n^{2}-1}=0$$ so the series is convergent based on the Cauchy's convergence test. But how do I calculate the sum? Thank you.

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marked as duplicate by Guy Fsone, Community Jan 16 '18 at 23:29

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ math.stackexchange.com/q/265277?rq=1 $\endgroup$ – Guy Fsone Jan 16 '18 at 23:20
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    $\begingroup$ Sorry I didn't realize I can put latex code in the search bar. I'm gonna do it next time. Thank you everyone for the answers! $\endgroup$ – bencemeszaros Jan 16 '18 at 23:32
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Hint. By a fraction decomposition, one gets $$ \frac{2}{4n^{2}-1}=\frac{1}{2n-1}-\frac{1}{2n+1} $$ then one may use a telescoping sum.

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Since$$\frac1{4n^2-1}=\frac1{(2n-1)(2n+1)}=\frac12\left(\frac1{2n-1}-\frac1{2n+1}\right),$$your series is a telescopic series.

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Your justification for the convergence to he series needs work (having a limit of zero of the summand does not imply the sum converges!), it does however converge a priori by noting that $$ \frac{1}{4n^2-1}=O\left(\frac{1}{n^2}\right) $$ So it converges by the $p$-test.

You may also use the partial fraction decomposition noted in the other answers and compute the telescoping series, showing that it converges.

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    $\begingroup$ Your equivalent is wrong as written. The $\sim$ has a specific meaning, and here you're missing the constant $4$ on the RHS. $\endgroup$ – Clement C. Jan 16 '18 at 23:14
  • $\begingroup$ i meant asymptotically equivalent to. what does it mean? $\endgroup$ – qbert Jan 16 '18 at 23:16
  • $\begingroup$ $a_n \sim_{n\to\infty} b_n$ if $\lim_{n\to\infty}\frac{a_n}{b_n} =1 $ (that's the simple definition; the actual one, equivalent when $b_n$ can't cancel, is that $a_n-b_n = o(b_n)$) $\endgroup$ – Clement C. Jan 16 '18 at 23:18
  • $\begingroup$ I see, I will amend the above then $\endgroup$ – qbert Jan 16 '18 at 23:19
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$$\frac{1}{4n^2-1}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$

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