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I'm working with the following sum and trying to determine what it converges to: $$\sum_{k=1}^{\infty}\ln\left(\frac{k(k+2)}{(k+1)^2}\right)$$

Numerically I see that it seems to be converging to $-\ln(2)$, however I can't see why that is the case. I have expanded the logarithm and expressed the sum as $\sum_{k=1}^{\infty}\ln(k)+\ln(k+2)-2\ln(k+1)$, but I don't know if that actually helps anything. Does anyone have any thoughts on how to approach this problem?

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marked as duplicate by Martin Sleziak, Arnaud D., Community Jan 8 '17 at 1:38

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  • $\begingroup$ Did you try to pull the sum into the logarithm? $\endgroup$ – flawr Apr 8 '16 at 21:30
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    $\begingroup$ Start writing out a few terms and see how they cancel out. It's a telescoping sum $\endgroup$ – David Quinn Apr 8 '16 at 21:30
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$$\frac{k^2+2k}{k^2+2k+1}=1-\frac1{(k+1)^2}\implies$$

$$\sum_{k=1}^n\log\left(1-\frac1{(k+1)^2}\right)=\sum_{k=1}^n\log\left[\left(1-\frac1{k+1}\right)\left(1+\frac1{k+1}\right)\right]=$$

$$=\sum_{k=1}^n\left[\log\left(1-\frac1{k+1}\right)+\log\left(1+\frac1{k+1}\right)\right]=$$

$$=\log\frac12+\overbrace{\log\frac32+\log\frac23}^{=\log1=0}+\overbrace{\log\frac43+\log\frac34}^{=\log1=0}+\log\frac54+\ldots+\log\frac n{n+1}+\log\frac{n+2}{n+1}=$$

$$=\log\frac12+\log\frac{n+2}{n+1}\xrightarrow[n\to\infty]{}-\log2$$

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$$ \begin{align} \sum_{k=1}^n\log\left(\frac{k(k+2)}{(k+1)^2}\right) &=\sum_{k=1}^n\log(k)+\sum_{k=1}^n\log(k+2)-2\sum_{k=1}^n\log(k+1)\\ &=\sum_{k=1}^n\log(k)+\sum_{k=3}^{n+2}\log(k)-2\sum_{k=2}^{n+1}\log(k)\\[3pt] &=\log(1)-\log(2)+\log(n+2)-\log(n+1)\\[6pt] &=-\log(2)+\log\left(\frac{n+2}{n+1}\right) \end{align} $$ Let $n\to\infty$, $$ \sum_{k=1}^\infty\log\left(\frac{k(k+2)}{(k+1)^2}\right)=-\log(2) $$

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In another way: $$ \eqalign{ & \sum\limits_{k\, = \,1}^n {\log \left( {{{k\left( {k + 2} \right)} \over {\left( {k + 1} \right)^2 }}} \right)} = \log \prod\limits_{k\, = \,1}^n {{{k\left( {k + 2} \right)} \over {\left( {k + 1} \right)^2 }}} = \cr & = \log \left( {\prod\limits_{k\, = \,1}^n {{k \over {\left( {k + 1} \right)}}} \;\prod\limits_{k\, = \,1}^n {{{\left( {k + 2} \right)} \over {\left( {k + 1} \right)}}} } \right) = ({\rm telescoping}) \cr & = \log \left( {{1 \over {\left( {n + 1} \right)}}\;{{n + 2} \over 2}} \right) \cr} $$

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  • $\begingroup$ Very nice indeed. +1 $\endgroup$ – DonAntonio Apr 8 '16 at 22:20

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