I have a sequence $(a_n)_{n\geq0}$ in $\mathbb{R}$ defined by $a_0 = 0$ and $a_n:=\sum_{k=1}^n \frac{(-1)^{k+1}}{k}$ for $n\geq1$.
I need to proof that $\lim_{n\to\infty} a_n$ exists.
I first try to proof $(a_{2n+1})_{n\geq0}$ is decreases monotonously, here I'm in doubt how to make the proof "closed". I currently have this, but I don't know if this actually proofs it.
$n+1>n$, then $a_{2(n+1)+1}=a_{2n+3}=a_{2n+1} +\sum_{2n+2}^{2n+3}=a_{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}<a_{2n+1}$ for all $n\in\mathbb{N}$. So $(a_{2n+1})_{n\geq0}$ montonously decreasing.
The proof for $(a_{2n})_{n\geq0}$ montonously increasing follows analogously.
I then think I have to proof that these are bounded below and bounded above respectively, and thereby convergent.
From there on I don't know what to do.
Yes, the sequence $(a_{2n-1})_{n\in\mathbb N}$ decreases and the sequence $(a_{2n})_{n\in\mathbb N}$ increases. Now, note that\begin{align}a_{2n-1}&=1-\frac12+\cdots+\frac1{2n-1}\\&>1-\frac12+\cdots+\frac1{2n-1}-\frac1{2n}\\&=a_{2n}\\&\geqslant a_2.\end{align}So, yes, the sequence $(a_{2n-1})_{n\in\mathbb N}$ is bounded. Since it is also monotonic, it converges. By the same argument, the sequence $(a_{2n})_{n\in\mathbb N}$ converges too. Now, it's all a matter of proving that the limits are the same. This is easy, since $a_{2n-1}-a_{2n}=\frac1{2n}\to0$.