Integral of a sum

Let $\displaystyle S = \sum_{n=1}^\infty \frac n{x^n}$ . Then we have:

$\begin{aligned} S & = \sum_{\color{#3D99F6}n=1}^\infty \frac n{x^n} = \sum_{\color{#D61F06}n=0}^\infty \frac n{x^n} \\ & = \sum_{\color{#D61F06}n=0}^\infty \frac {n+1}{x^{n+1}} \\ & = \frac 1x \left(\sum_{n=0}^\infty \frac n{x^n} + \sum_{n=0}^\infty \frac 1{x^n} \right) \\ & = \frac 1x \left(S + \frac 1{1-\frac 1x} \right) \\ xS & = S + \frac x{x-1} \end{aligned}$

$\implies S = \dfrac x{(x-1)^2}$

Therefore,

$\begin{aligned} I & = \int_{m+1}^{m+2} \sum_{n=1}^\infty \frac n{x^n} dx \\ & = \int_{m+1}^{m+2} \frac x{(x-1)^2} dx \\ & = \int_{m+1}^{m+2} \frac {x-1+1}{(x-1)^2} dx \\ & = \int_{m+1}^{m+2} \left(\frac 1{x-1} + \frac 1{(x-1)^2} \right) dx \\ & = \ln (x-1) - \frac 1{x-1} \bigg|_{m+1}^{m+2} \\ & = \ln \left(\frac {m+1}m\right) - \frac 1{m+1} + \frac 1m \\ & = \ln \left(\frac {m+1}m\right) + \frac 1{m(m+1)} \\ & = \ln \left(\frac {m+1}m\right) + \frac {{\color{#3D99F6}m^2+m}+1}{m(m+1)} - \frac {\color{#D61F06}m^2+m}{m(m+1)} \\ & = {\color{#D61F06}\ln \left(\frac {m+1}m\right)} + \frac {m^2+m+1}{m(m+1)} \color{#D61F06}-1 \end{aligned}$

This implies that:

$\begin{aligned} \ln \left(\frac {m+1}m \right) & = 1 \\\frac {m+1}m & = e \\ \implies m & = \frac 1{e-1} \approx \boxed{0.5819767} \end{aligned}$

$S(x) = \sum_{n = 1}^{\infty} \dfrac{n}{x^n} = \dfrac{1}{x}(1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} + ... ) + \dfrac{1}{x^2}(1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} + ...) +\dfrac{1}{x^3}(1 + \dfrac{1}{x} + \dfrac{1}{x^2} + \dfrac{1}{x^3} + ...) + ... =$

$(\dfrac{x}{x - 1})(\dfrac{1}{x})(\dfrac{x}{x - 1}) = \dfrac{x}{(x - 1)^2} = \dfrac{x - 1 + 1}{(x - 1)^2} = \dfrac{1}{x - 1} + \dfrac{1}{(x - 1)^2}$ .

$\int_{m + 1}^{m + 2} S(x) dx = (\ln(x - 1) - \dfrac{1}{x - 1})|_{m + 1}^{m + 2} = \ln(\dfrac{m + 1}{m}) + \dfrac{1}{m(m + 1)} = \dfrac{m^2 + m + 1}{m(m + 1)} = 1 + \dfrac{1}{m(m + 1)} \implies$

$\ln(\dfrac{m + 1}{m}) = 1 \implies \dfrac{m + 1}{m} = e \implies m = \dfrac{1}{e - 1} = \boxed{0.5819767}$ to seven decimal places.