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How do I evaluate the sum

$$\sum_{k=1}^\infty\left(\ln\Big(1+\frac{1}{k+a} \Big)-\ln\Big(1+\frac{1}{k+b}\Big)\right)$$

where $0 <a<b<1$?

Hints will be appreciated

Thanks

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closed as off-topic by Did, Claude Leibovici, colormegone, user228113, Shailesh Jan 9 '16 at 5:13

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  • 3
    $\begingroup$ What are $a$ and $b$? $\endgroup$ – luka5z Jan 1 '16 at 9:06
  • $\begingroup$ @luka5z edited thanks $\endgroup$ – Taylor Ted Jan 1 '16 at 9:16
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We have $$ \sum_{k=1}^n \ln\left(1+\frac{1}{k+a}\right)=\ln\left(\prod_{k=1}^n\Big(1+\frac{1}{k+a}\Big)\right)=\ln\left(\prod_{k=1}^n\frac{k+1+a}{k+a}\right)=\ln\left(\frac{n+1+a}{1+a}\right) $$ and thus $$ \sum_{k=1}^n \ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)= \ln\left(\frac{n+1+a}{1+a}\right)-\ln\left(\frac{n+1+b}{1+b}\right) \\=\ln\left(\frac{1+b}{1+a}\right)-\ln\left(\frac{n+1+b}{n+1+a}\right)\to \ln\left(\frac{1+b}{1+a}\right), $$ as $n\to\infty$, since $$ \ln\left(\frac{n+1+b}{n+1+a}\right)=\ln\left(1+\frac{b-a}{n+1+a}\right)\to 0. $$

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  • $\begingroup$ But answer is not 0. $\endgroup$ – Taylor Ted Jan 1 '16 at 9:37
  • $\begingroup$ @Fabian You are right. I corrected it. $\endgroup$ – Yiorgos S. Smyrlis Jan 1 '16 at 9:44
  • $\begingroup$ @TaylorTed I didn't get $0$ as an answer, but $\ln(\frac{1+b}{1+a})$. $\endgroup$ – Yiorgos S. Smyrlis Jan 1 '16 at 9:45
  • $\begingroup$ @YiorgosS.Smyrlis i am not getting how summation is changed into product $\endgroup$ – Taylor Ted Jan 1 '16 at 9:49
  • $\begingroup$ @TaylorTed Since $\sum \ln a_i=\ln a_1+\cdots +\ln a_n=\ln (a_1\cdots a_n)=\ln (\prod a_i)$. $\endgroup$ – Yiorgos S. Smyrlis Jan 1 '16 at 9:51
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Let $$S_n = \ln\prod\limits_{k=1}^n\dfrac{k+a+1}{k+a} - \ln\prod\limits_{k=1}^n\dfrac{k+b+1}{k+b}= \ln\dfrac{k+a+1}{a+1} - \ln\dfrac{k+b+1}{b+1}$$$$ = \ln\dfrac{b+1}{a+1} + \ln\dfrac{k+b+1}{k+a+1}.$$ $$\sum_{k=1}^{\infty} \ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right) = \lim\limits_{n\to \infty} S_n = \ln\dfrac{b+1}{a+1} +\ln1 =\ln\dfrac{b+1}{a+1}$$

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Note that, for each fixed $k$, $\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)=\ln\left(\dfrac{1+\frac{1}{k+a}}{1+\frac{1}{k+b}}\right)$. Then:

\begin{eqnarray*} \sum_{k=1}^n\ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)&=&\sum_{k=1}^n\ln\left(\dfrac{1+\frac{1}{k+a}}{1+\frac{1}{k+b}}\right)\\ &=&\ln\left(\prod_{k=1}^n\dfrac{1+\frac{1}{k+a}}{1+\frac{1}{k+b}}\right)\\ &=&\ln\left(\dfrac{\frac{n+a+1}{a+1}}{\frac{n+b+1}{b+1}}\right)\\ &=&\ln\left(\dfrac{1+\frac{n}{a+1}}{1+\frac{n}{b+1}}\right)\\ &=&\ln\left(\dfrac{\frac{1}{n}+\frac{1}{a+1}}{\frac{1}{n}+\frac{1}{b+1}}\right) \end{eqnarray*}

Thus, taking $n\to\infty$, the limit is $\ln(b+1)-\ln(a+1)$

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  • $\begingroup$ Why summation is changed into product $\endgroup$ – Taylor Ted Jan 1 '16 at 9:39
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    $\begingroup$ Properties of log of a product: $\ln(xy)=\ln(x)+\ln(y)$, where $x,y>0$, and it extends to a finite number of summands (that´s the reason for we take first the sum from $k=1$ to $n$ and finally take limit) $\endgroup$ – sinbadh Jan 1 '16 at 9:41

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