<s>[INST] 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?
 [/INST] 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. :/
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