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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https://math.stackexchange.com/q/265277?rq=1 – Guy Fsone Jan 16 '18 at 23:20
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2Sorry 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! – 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.
Olivier Oloa
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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.
José Carlos Santos
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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.
operatorerror
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2Your equivalent is wrong as written. The $\sim$ has a specific meaning, and here you're missing the constant $4$ on the RHS. – Clement C. Jan 16 '18 at 23:14
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$a_n \sim_{n\to\infty} b_n$ 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 $a_n-b_n = o(b_n)$) – Clement C. Jan 16 '18 at 23:18
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