Can Riemann integral lemma be helpful? Let $0<\alpha<\beta<1.$  Compute $$
\sum_{k=1}^{\infty} \int_{1 /(k+\beta)}^{1 /(k+\alpha)} \frac{1}{1+x}\,\mathrm{d}x$$
My approach:- putting the limit into $\ln(1+x)$ ,I am getting $$\ln\frac{(k+\alpha+1)(k+\beta)}{(k+\beta+1)(k+\alpha)}$$
what  to do next? 
 A: Let $0<\alpha<\beta<1.$  We have to compute $$\sum_{k=1}^{\infty} \int_{1 /(k+\beta)}^{1 /(k+\alpha)} \frac{1}{1+x}\,\mathrm{d}x.$$
For all $n\in\mathbb{N}$ we get
$\sum_{k=1}^n \int_{1 /(k+\beta)}^{1 /(k+\alpha)} \frac{1}{1+x}\,\mathrm{d}x=$
$=\sum_{k=1}^n\ln\frac{(k+1+\alpha)(k+\beta)}{(k+1+\beta)(k+\alpha)}=$
$=\ln\frac{(2+\alpha)(1+\beta)}{(2+\beta)(1+\alpha)}+ \ln\frac{(3+\alpha)(2+\beta)}{(3+\beta)(2+\alpha)}+  \ln\frac{(4+\alpha)(3+\beta)}{(4+\beta)(3+\alpha)}+...+\\\;\;+\ln\frac{(n+1+\alpha)(n+\beta)}{(n+1+\beta)(n+\alpha)}=$
$=\ln\frac{(2+\alpha)(1+\beta)(3+\alpha)(2+\beta) (4+\alpha)(3+\beta)\cdot...\cdot(n+1+\alpha)(n+\beta)}{(2+\beta)(1+\alpha)(3+\beta)(2+\alpha)(4+\beta)(3+\alpha)\cdot...\cdot(n+1+\beta)(n+\alpha)}=$
$=\ln\frac{(1+\beta)(n+1+\alpha)}{(1+\alpha)(n+1+\beta)}.$
Hence
$$\sum_{k=1}^{\infty} \int_{1 /(k+\beta)}^{1 /(k+\alpha)} \frac{1}{1+x}\,\mathrm{d}x=\lim_{n\rightarrow\infty} \sum_{k=1}^n \int_{1 /(k+\beta)}^{1 /(k+\alpha)} \frac{1}{1+x}\,\mathrm{d}x=\\=\lim_{n\rightarrow\infty} \ln\frac{(1+\beta)(n+1+\alpha)}{(1+\alpha)(n+1+\beta)}= \ln\frac{1+\beta}{1+\alpha}$$
