$$$\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k \left(k + 2\right)}$$$

$$$\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k \left(k + 2\right)}$$$ を求めよ。

Perform partial fraction decomposition (for steps, see partial fraction decomposition calculator)::

$${\color{red}{\left(\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k \left(k + 2\right)}\right)}}={\color{red}{\left(\sum_{k=1}^{\infty} \left(- \frac{\left(-1\right)^{k}}{2 \left(k + 2\right)} + \frac{\left(-1\right)^{k}}{2 k}\right)\right)}}$$

Split the series:

$${\color{red}{\left(\sum_{k=1}^{\infty} \left(- \frac{\left(-1\right)^{k}}{2 \left(k + 2\right)} + \frac{\left(-1\right)^{k}}{2 k}\right)\right)}}={\color{red}{\left(\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{2 k} + \sum_{k=1}^{\infty} - \frac{\left(-1\right)^{k}}{2 \left(k + 2\right)}\right)}}$$

Pull the constant out of the series:

$${\color{red}{\left(\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{2 k}\right)}} + \sum_{k=1}^{\infty} - \frac{\left(-1\right)^{k}}{2 \left(k + 2\right)}={\color{red}{\left(\frac{\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k}}{2}\right)}} + \sum_{k=1}^{\infty} - \frac{\left(-1\right)^{k}}{2 \left(k + 2\right)}$$

$$$\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k}$$$ is a known series.

It is $$$\sum_{k=1}^{\infty} \left(-1\right)^{k} k^{- n_{0}}=- \eta\left(n_{0}\right)$$$, $$$n_{0} > 0$$$ with $$$n_{0}=1$$$.

Therefore,

$$\frac{{\color{red}{\left(\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k}\right)}}}{2} + \sum_{k=1}^{\infty} - \frac{\left(-1\right)^{k}}{2 \left(k + 2\right)}=\frac{{\color{red}{\left(- \ln{\left(2 \right)}\right)}}}{2} + \sum_{k=1}^{\infty} - \frac{\left(-1\right)^{k}}{2 \left(k + 2\right)}$$

Rewrite:

$$- \frac{\ln{\left(2 \right)}}{2} + {\color{red}{\left(\sum_{k=1}^{\infty} - \frac{\left(-1\right)^{k}}{2 \left(k + 2\right)}\right)}}=- \frac{\ln{\left(2 \right)}}{2} + {\color{red}{\left(\sum_{k=1}^{\infty} - \frac{\left(-1\right)^{k + 2}}{2 \left(k + 2\right)}\right)}}$$

Shift the series by $$$2$$$:

$$- \frac{\ln{\left(2 \right)}}{2} + {\color{red}{\left(\sum_{k=1}^{\infty} - \frac{\left(-1\right)^{k + 2}}{2 \left(k + 2\right)}\right)}}=- \frac{\ln{\left(2 \right)}}{2} + {\color{red}{\left(\sum_{k=3}^{\infty} - \frac{\left(-1\right)^{k}}{2 k}\right)}}$$

Pull the constant out of the series:

$$- \frac{\ln{\left(2 \right)}}{2} + {\color{red}{\left(\sum_{k=3}^{\infty} - \frac{\left(-1\right)^{k}}{2 k}\right)}}=- \frac{\ln{\left(2 \right)}}{2} + {\color{red}{\left(- \frac{\sum_{k=3}^{\infty} \frac{\left(-1\right)^{k}}{k}}{2}\right)}}$$

Split the series:

$$- \frac{{\color{red}{\left(\sum_{k=3}^{\infty} \frac{\left(-1\right)^{k}}{k}\right)}}}{2} - \frac{\ln{\left(2 \right)}}{2}=- \frac{{\color{red}{\left(\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k} + \sum_{k=1}^{2} - \frac{\left(-1\right)^{k}}{k}\right)}}}{2} - \frac{\ln{\left(2 \right)}}{2}$$

Since the bounds are finite, the number of terms is finite as well, and we just calculate the sum by summing up the terms.

$$- \frac{{\color{red}{\left(\sum_{k=1}^{2} - \frac{\left(-1\right)^{k}}{k}\right)}}}{2} - \frac{\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k}}{2} - \frac{\ln{\left(2 \right)}}{2}=- \frac{{\color{red}{\left(\frac{1}{2}\right)}}}{2} - \frac{\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k}}{2} - \frac{\ln{\left(2 \right)}}{2}$$

$$$\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k}$$$ is a known series.

It is $$$\sum_{k=1}^{\infty} \left(-1\right)^{k} k^{- n_{0}}=- \eta\left(n_{0}\right)$$$, $$$n_{0} > 0$$$ with $$$n_{0}=1$$$.

Therefore,

$$- \frac{1}{4} - \frac{{\color{red}{\left(\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k}\right)}}}{2} - \frac{\ln{\left(2 \right)}}{2}=- \frac{\ln{\left(2 \right)}}{2} - \frac{1}{4} - \frac{{\color{red}{\left(- \ln{\left(2 \right)}\right)}}}{2}$$

Hence,

$$\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k \left(k + 2\right)}=- \frac{1}{4}$$

$$$\sum_{k=1}^{\infty} \frac{\left(-1\right)^{k}}{k \left(k + 2\right)} = - \frac{1}{4} = -0.25$$$A