Prove that $a_{2n+1}$ and $a_{2n}$ are convergent, and that $a_{2n+1}-a_{2n} \to 0$ when $n \to \infty$.

Question

For each $n \geq 1$ we define $a_n= \sum_{k=1}^{n} \frac{(-1)^{k+1}}{k}$.

Prove that $a_{2n+1}$ and $a_{2n}$ are convergent, and that $a_{2n+1}-a_{2n} \to 0$ when $n \to \infty$.

So far I managed to prove that $a_{2n+1}$ is monotone decreasing, and $a_{2n}$ is monotone increasing.

I wanted to use that the sum is bounded, and since $a_{2n+1}$ is monotone decreasing, and $a_{2n}$ is monotone increasing they would be convergent. I proved that $a_{2n+1}$ is bounded below by $\frac{1}{2}$. I'm having trouble proving $a_{2n}$ is bounded above.

Answer 1

All you need to do is to notice that $(\forall n\in\mathbb{N}):a_{2n}<a_{2n+1}\leqslant a_1$.

Answer 2

$a_{2n+1} = a_{2n} + \dfrac{1}{2n+1} $. I hope this hint helps.