<s>[INST] Prove the sequence of general term $\sum\limits_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k}$ converges 
Given the sequence $$a_n :=\sum_{k=1}^n \frac{(-1)^{k+1}}{k}$$
  Prove the sequence $(a_{2n+1})_{n \geq 0}$ converges. 

My thoughts
I've proven that the sequence $(a_{2n+1})_{n \geq 0}$ is monotone decreasing. So now I want to prove that the sequence is bounded below, since then I can prove the sequence converges.
I wanted to prove the sequence is bounded below by induction, however I'm not quite sure how.
Since $$a_{2n+1} :=\sum_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k}= 1 -\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$$
I think it is bounded by $\frac{1}{2}$. 
Using induction:


*

*For $n=0$, we know $a_{2n+1} = a_1 = 1 \geq\frac{1}{2}$

*Suppose that $a_{2n+1} \geq \frac{1}{2}$. Prove that $a_{2(n+1)+1)} = a_{2n+3} \geq \frac{1}{2}$.
I know that $$a_{2n+3} :=\sum_{k=1}^{2n+3} \frac{(-1)^{k+1}}{k} = a_{2n+1} -\frac{1}{2n+2} + \frac{1}{2n+3} = a_{2n+1} -\frac{1}{(2n+2)(2n+3)}$$ But I don't know how to proceed. Any help?

 [/INST] To show that $a_{2n+1} > \frac12$, note that
$$
a_{2n+1} = \sum_{k=1}^{2n+1} \frac{(-1)^{k+1}}{k} = \underbrace{1 - \frac12}_{=\frac12} + \underbrace{\frac13 - \frac14}_{>0} + \underbrace{\frac15 - \frac16}_{>0} + \ldots + \underbrace{\frac{1}{2n-1} - \frac{1}{2n}}_{>0} + \underbrace{\frac{1}{2n+1}}_{>0} > \frac12.
$$
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