Let $a(n)=\sum_{r=1}^{n}(-1)^{r-1}\frac{1}{r}$. Prove that $a(2n) \neq 1$ for any value $n$. Let $$a(n)=\sum_{r=1}^{n}(-1)^{r-1}\frac{1}{r}.$$
I experienced this function while doing a problem, I could do the problem, but I got stuck at a point where I had to prove that $a(2n)<1$ for all $n$. I proved that $a(2n) \le 1$ for all $n$ so in order to prove what the question demands, I need to prove that $a(2n) \neq 1$ for any value $n$. Nothing striked my mind as of now how to prove it.
So can someone help me proving that $a(2n) \neq 1$ for any value  $n$?
 A: One way is to pair up adjacent terms (since $2n$ is even every term has a buddy) in order to write
$$
a(2n)
= \sum_{r = 1}^{2n} (-1)^{r - 1} \frac{1}{r}
= \sum_{r = 1}^n \left(\frac{1}{2 r} - \frac{1}{2 r + 1}\right)
= \sum_{r = 1}^n \frac{1}{2 r (2 r + 1)}
\leq \sum_{r = 1}^n \frac{1}{(2 r)^2}
\leq \frac{1}{4} \sum_{r = 1}^\infty \frac{1}{r^2}
= \frac{1}{4} \frac{\pi^2}{6} < 1.
$$
A: Here's a solution that leverages the identity
$$\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}=\sum_{r=1}^{2n}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r}$$
I'd suggest you try and prove this yourself. If that's not an option, I've provided a proof below:

 Proof:\begin{align} \sum_{r=1}^{2n}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r} &= \left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2n-1}+\frac{1}{2n}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n-1}+\frac{1}{n}\right)\\&= 1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{1}{2}\right)+\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{3}\right)+\cdots+\frac{1}{2n-1}+\left(\frac{1}{2n}-\frac{1}{n}\right)\\&= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}\\&=\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}\end{align}

Armed with this identity, we can write
\begin{align}
\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r} &= \sum_{r=1}^{2n}\frac{1}{r}-\sum_{r=1}^{n}\frac{1}{r}\\
&= \sum_{r=n+1}^{2n}\frac{1}{r}\\
&\leq \sum_{r=n+1}^{2n}\frac{1}{n+1}\\
&= \frac{2n-(n+1)+1}{n+1}\\
&= \frac{n}{n+1}
\end{align}
so $\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}\leq\frac{n}{n+1}$. Since $\frac{n}{n+1}<\frac{n+1}{n+1}=1$ for every $n\geq 1$, this immediately gives the desired result.
$$a(2n)=\sum_{r=1}^{2n}\frac{(-1)^{r-1}}{r}<1$$
A: $\frac{1}{n}$ is strictly decreasing, so $a(2n)$ is strictly increasing. $$\lim_{n\to\infty}a(2n)=\ln 2,$$ so
$$0\leq a(2n)<\ln 2< 1$$
