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I have to show that the series $\sum_{n=1}^{\infty} \log\left(\frac{(2n)^2}{(2n+1)(2n-1)}\right)$ converges.

I have tried Ratio Test and Cauchy Condensation Test but it didn't work for me. I tried using Comparison Test but I couldn't make an appropriate inequality for it. Could you please give me some hints. Any help will be appreciated.

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  • $\begingroup$ Hint: Split up the logarithms and pull a $\log((2n)^2)$ out of the second part. U will end up with something like- $\sum_n \log(1+1/(4n^2))$ can u conclude? $\endgroup$ – tired Oct 14 '15 at 17:33
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    $\begingroup$ Fun fact: The sum equals -$\log[2/\pi]$ $\endgroup$ – tired Oct 14 '15 at 17:45
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$0<\log (1+x)<x$ for $x>0 . $ Therefore $0<\log (\frac{4 n^2}{4 n^2-1})=\log (1+\frac{1}{4 n^2-1})< \frac{1}{4 n^2-1} < \frac{1}{2 n^2}. $ The sum $\sum (1/2 n^2)$ converges by Cauchy Condensation. Your sum therefore converges by Comparison.

(Where did that term $\frac{1}{2 n^2}$ come from in the inequality? From the idea, with $4 n^2=x$, that $\frac{1}{x-1}<\frac{2}{x}$ if $x$ is big enough.)

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  • $\begingroup$ This is concise and effective! +1 $\endgroup$ – Mark Viola Oct 15 '15 at 1:03
  • $\begingroup$ @Dr MV .Thank you...................... $\endgroup$ – DanielWainfleet Oct 15 '15 at 2:38
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I couldn't resist

We use the fact that $e^{\sum\log (s_n)}=\prod s_n $

Setting $s_n=\frac{1}{1-\frac{1}{4n^2}}$, we find that

$$ e^{\sum_{n=1}^{\infty}\log (s_n)}= \prod_{n=1}^{\infty} \left(1-\frac{1}{4n^2}\right)^{-1} $$

Using the product expansion of $\sin(\pi x)$

$$ \frac{\pi x}{\sin(\pi x)}=\frac{1}{\prod_{n=1}^{\infty} \left(1-\frac{x^2}{n^2}\right)} $$

and setting $x=1/2$ we immediately find that

$$ e^{\sum_{n=1}^{\infty} \log(s_n)}=\frac{\pi}{2} $$

and therefore

$$ \sum_{n=1}^{\infty}\log\left(\frac{1}{1-\frac{1}{4n^2}}\right)=\log\left(\frac{\pi}{2}\right) $$

And therefore the sum converges ;)

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    $\begingroup$ Show-off.You beat me to it. $\endgroup$ – DanielWainfleet Oct 14 '15 at 19:00
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    $\begingroup$ Nice Solution! +1 $\endgroup$ – Mark Viola Oct 15 '15 at 1:01
  • $\begingroup$ I don't think somebody who came up with such a question would understand your solution. It is quoted that "while it is great to respect the intelligence of your reader, you should not overtax him either." $\endgroup$ – user98186 Jan 3 '16 at 17:31
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$$\frac{4 n^2}{4 n^2-1} = \frac1{1-\frac1{4 n^2}}$$

Thus,

$$\log{\left (\frac{4 n^2}{4 n^2-1} \right )} = -\log{\left ( 1-\frac1{4 n^2} \right )}$$

which, for large $n$, is approximately equal to $1/(4 n^2)$. Thus the sum converges by the comparison test with $\zeta(2)$.

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Since $\log (1 + x) < x$ for non-negative $x$, we have

$$\begin {eqnarray} 0 < \sum_{n = 1}^{\infty} \log \left (\frac {(2n)^2} {(2n - 1) (2n + 1)} \right) & = & \sum_{n = 1}^{\infty} \log \left (1 + \frac {1} {(2n - 1) (2n + 1)} \right) \nonumber \\ & < & \sum_{n = 1}^{\infty} \left ( \frac {1} {(2n - 1) (2n + 1)} \right) \nonumber \\ & = & \frac {1} {2} \left (\sum_{n = 1}^{\infty} \frac {1} {2n - 1} - \sum_{n = 1}^{\infty} \frac {1} {2n + 1} \right) \nonumber \\ & = & \frac {1} {2} \left (\sum_{n = 0}^{\infty} \frac {1} {2n + 1} - \sum_{n = 1}^{\infty} \frac {1} {2n + 1} \right) \nonumber \\ & = & \frac {1} {2} \cdot 1 = \frac {1} {2}. \end {eqnarray}$$

So, the series converges.

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