Encontrando el valor de esta suma.

Question

Queremos encontrar el valor de esta serie $\sum_{n=1}^{\infty} \frac{n}{2^n(n+1)}$. Según Wolfram, el valor es $2 - ln(4)$

Lo que hice:

$\frac{1}{1-x}$ = $\sum_{n=0}^{\infty} x^n$. Diferenciación:

$\frac{1}{(1-x)^2}$ = $\sum_{n=1}^{\infty} nx^{n-1}$

$\frac{x}{(1-x)^2}$ = $\sum_{n=1}^{\infty} nx^{n}$. Integración:

$\frac{1}{(1-x)}+ ln(\vert x-1 \vert) $ = $\sum_{n=1}^{\infty} \frac{nx^{n+1}}{n+1}$

$\frac{1}{x(1-x)} + \frac{ln(\vert x-1 \vert)}{x} $ = $\sum_{n=1}^{\infty} \frac{nx^{n}}{n+1}$

Haciendo $x = \frac{1}{2}$, tenemos la serie deseada. Pero la función en $x= \frac{1}{2}$ es $2,61…$, no $2 - ln(4)$

¿Qué estoy haciendo mal?

Answer 1

Para todo $a > 1$,

$$ \begin{align} \sum_{k \ge 1} \frac {k}{k + 1} \cdot \frac {1}{{a}^{k}} & = \sum_{k \ge 1} \frac {k}{{a}^{k}} \int_{0}^{1} {x}^{k} \text {d} x \\ & = \int_{0}^{1} \sum_{k \ge 1} k \, {\Big( \frac {x}{a} \Big)}^{k} \text {d} x \\ & = \int_{0}^{1} \sum_{k \ge 1} {\bigg[ t \cdot \frac {\text {d}}{\text {d} t} {t}^{k} \bigg]}_{t = \frac {x}{a}} \text {d} x \\ & = \int_{0}^{1} {\bigg[ t \cdot \frac {\text {d}}{\text {d} t} \sum_{k \ge 1} {t}^{k} \bigg]}_{t = \frac {x}{a}} \text {d} x \\ & = \int_{0}^{1} {\bigg[ t \cdot \frac {\text {d}}{\text {d} t} \frac {t}{1 - t} \bigg]}_{t = \frac {x}{a}} \text {d} x \\ & = \int_{0}^{1} {\bigg[ t \cdot \frac {\text {d}}{\text {d} t} \Big( \frac {1}{1 - t} - 1 \Big) \bigg]}_{t = \frac {x}{a}} \text {d} x \\ & = \int_{0}^{1} {\Bigg[ t \cdot \frac {1}{{\left( 1 - t \right)}^{2}} \Bigg]}_{t = \frac {x}{a}} \text {d} x \\ & = a \, \int_{0}^{\frac {1}{a}} \frac {x}{{\left( 1 - x \right)}^{2}} \text {d} x \\ & = a \, \int_{1 - \frac {1}{a}}^{1} \frac {1 - x}{{x}^{2}} \text {d} x \\ & = a \, \int_{1}^{\frac {a}{a - 1}} \Big( 1 - \frac {1}{x} \Big) \text {d} x \\ & = \frac {a}{a - 1} - a \, \ln \Big( \frac {a}{a - 1} \Big). \end{align} $$

En tu caso, tenemos $a = 2$.