<s>[INST] Convergence of $\sum\limits_{n=1}^{\infty}\frac{1+(-1)^n}{n}$ I want to check, whether $\sum\limits_{n=1}^{\infty}\frac{1+(-1)^n}{n}$ converges or diverges.
Leibniz's test failed, and ratio test just made it even more complicated, so i tried to use the comparison test, but i can't find a suitable series so that $\lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n}$ exists..
 [/INST] To prove: $\sum_{n\geq 1} \frac{1+(-1)^n}{n}$ diverges.
Proof: \begin{align*} \sum _{n\geq 1} \frac{1+(-1)^n}{n} &= \sum _{k\geq 1} \frac{1+(-1)^{2k}}{2k} + \sum _{k\geq 1} \frac{1+(-1)^{2k-1}}{2k-1} \\
&= \sum _{k\geq 1} \frac{2}{2k} + \sum _{k\geq 1} \frac{0}{2k-1} \\\
&= \sum _{k\geq 1} \frac{1}{k} \end{align*}
Because $\sum _{k\geq 1} \frac{1}{k}$ diverges, $\sum_{n\geq 1} \frac{1+(-1)^n}{n}$ diverges as well. \qed
Improved answer (with partial sums):
For each $m\in \mathbb{N}$ we have
\begin{align*}
\sum _{n=1} ^m \frac{1 + (-1)^n}{n} 
& = \sum _{k=1} ^{\lfloor m/2 \rfloor} \frac{1 + (-1)^{2k}}{2k} + \sum _{k=1} ^{\lfloor \frac{m+1}{2}\rfloor} \frac{1 + (-1)^{2k-1}}{2k-1} \\
& = \sum _{k=1} ^{\lfloor m/2 \rfloor} \frac{2}{2k} + \sum _{k=1} ^{\lfloor \frac{m+1}{2}\rfloor} \frac{0}{2k-1} 
= \sum _{k=1} ^{\lfloor m/2 \rfloor} \frac{1}{k}
\end{align*}
Because $\sum _k 1/k $ diverges, $\sum _n \frac{1 + (-1)^n}{n}$ diverges as well. QED
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