How do I evaluate the sum $\sum_{k=1}^\infty\left(\ln\big(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\big)\right)$

Question

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

Answer 1

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 $$ \ln\left(\frac{n+1+b}{n+1+a}\right)=\ln\left(1+\frac{b-a}{n+1+a}\right)\to 0. $$

Answer 2

Let $$S_n = \ln\prod\limits_{k=1}^n\dfrac{k+a+1}{k+a} - \ln\prod\limits_{k=1}^n\dfrac{k+b+1}{k+b}= \ln\dfrac{k+a+1}{a+1} - \ln\dfrac{k+b+1}{b+1}$$$$ = \ln\dfrac{b+1}{a+1} + \ln\dfrac{k+b+1}{k+a+1}.$$ $$\sum_{k=1}^{\infty} \ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right) = \lim\limits_{n\to \infty} S_n = \ln\dfrac{b+1}{a+1} +\ln1 =\ln\dfrac{b+1}{a+1}$$

Answer 3

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 $\ln(b+1)-\ln(a+1)$