I have to show that the sum of this double series is $\frac{1}{2}$ i have to solve this double series. i tried it, but i am not sure, that it is enough.
$$\sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \left(\left(\frac{1}{k+1} \cdot \left(\frac{k}{k+1}\right)^{i}\right) - \left(\frac{1}{k+2} \cdot \left(\frac{k+1}{k+2}\right)^{i}\right)\right)$$
$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{(1)^{i}}{k+1}  - \frac{(1)^{i}}{k+2} $$
$$= \sum_{i=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{k+1}  - \frac{1}{k+2} $$
this is a telescoping series. Because of that, the Series is convergence to $\lim_{n\to\infty} \frac{1}{2} - \frac{1}{n+2} = \frac{1}{2}$
Is this solution right and enough?
 A: Given that
$$\color{red}{\sum_{i=1}^{\infty}} \sum_{k=1}^{\infty} \bigg(\bigg(\frac{1}{k+1} \cdot \bigg(\frac{k}{k+1}\bigg)^{i}\bigg) - \bigg(\frac{1}{k+2} \cdot \bigg(\frac{k+1}{k+2}\bigg)^{i}\bigg)\bigg)$$
let's deal with the black sum first.
$$\begin{align}
&\sum_{k=1}^{\infty} \bigg(\underbrace{\bigg(\frac{1}{k+1} \cdot \bigg(\frac{k}{k+1}\bigg)^{i}}_{a_k}\bigg) - \bigg(\underbrace{\frac{1}{k+2} \cdot \bigg(\frac{k+1}{k+2}\bigg)^{i}\bigg)}_{a_{k+1}}\bigg)  \\
&= \lim_{n\to\infty}\sum_{k=1}^n(a_k - a_{k+1}) \\
&= \lim_{n\to\infty}(a_1 - a_{n+1}) \\
&= \lim_{n\to\infty}\left(\frac{1}{2}\left(\frac{1}{2}\right)^i - \left(\frac{1}{n+2}\left(\frac{n+1}{n+2}\right)^i\right)\right) \\
&= \left(\frac{1}{2}\right)^{i+1}
\end{align}$$
Now, we will move to the red sum:
$$\color{red}{\sum_{i=1}^\infty} \left(\frac{1}{2}\right)^{i+1} = \frac{\left(\frac{1}{2}\right)^2}{1 - \frac{1}{2}} = \frac{1}{4}\cdot 2 = \frac{1}{2}$$
A: Here is my attempt.
Since all terms are positive, we are allowed to swap the two summations.
$$\sum_{i=1}^\infty \sum_{k=1}^{\infty} \left( \frac{1}{k+1} (\frac{k}{k+1})^i - \frac{1}{k+2} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty }\sum_{k=1}^N \sum_{i=1}^{\infty} \left( \frac{1}{k+1} (\frac{k}{k+1})^i - \frac{1}{k+2} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{1}{k+1} \sum_{i=1}^{\infty} (\frac{k}{k+1})^i - \sum_{k=1}^N \frac{1}{k+2} \sum_{i=1}^{\infty} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{1}{k+1} \sum_{i=1}^{\infty} (\frac{k}{k+1})^i - \sum_{k=1}^N \frac{1}{k+2} \sum_{i=1}^{\infty} (\frac{k+1}{k+2})^i \right) = $$
$$= \lim_{N \to \infty } \left( \sum_{k=1}^N \frac{k}{k+1}  - \sum_{k=1}^N \frac{k+1}{k+2}  \right) = $$
$$= \lim_{N \to \infty } \left( \frac{1}{2} - \frac{N+1}{N+2}  \right) = - \frac{1}{2}$$
I don't know where I did a minus sign mistake. :/
