TITLE: Proof verification: $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$
QUESTION [18 upvotes]: Motivated by the beautiful identity
$$\int_{0}^{\frac{\pi}{4}}\tan^{2n}(x)\text{ }dx=(-1)^n\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)$$
and the equally beautiful proof of
$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=\frac{\pi}{4}$$
that it implies (the proof is presented in this post), I decided to try and find a closed-form expression for $\int_{0}^{\pi/4}\tan^{2n+1}(x)dx$. Following the method I used to establish the integral above (reduction formulae), I discovered this stunning identity, true for all $n\geq 1$ (also true for $n=0$ if you accept the empty sum convention).
$$\int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx=(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)$$
As you can probably see, this is very suggestive of the fact that $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$, so I naturally tried to leverage it in a proof. Here's my attempt:
Proof: We'll begin by proving that $\int_{0}^{\pi/4}\tan^{2n+1}(x)\text{ }dx=(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)$. We do this with mathematical induction.
The base case $n=1$ can be established as follows:
\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan^{2(1)+1}(x)dx &= \int_{0}^{\frac{\pi}{4}}\tan^{3}(x)\text{ }dx\\
&= \int_{0}^{\frac{\pi}{4}}\tan(x)\left[\sec^2(x)-1\right]dx\\
&= \int_{0}^{\frac{\pi}{4}}\tan(x)\sec^2(x)\text{ }dx-\int_{0}^{\frac{\pi}{4}}\tan(x)\text{ }dx
\end{align*}
Making the substitution $u=\tan(x)$ in the first integral gives
\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan(x)\sec^2(x)\text{ }dx-\int_{0}^{\frac{\pi}{4}}\tan(x)\text{ }dx &= \int_{0}^{1}u\text{ }du-\left[\ln\left(\sec\frac{\pi}{4}\right)-\ln(\sec 0)\right]\\
&= \frac{1}{2}-\ln(\sqrt{2})\\
&= -\left(\frac{1}{2}\ln(2)-\frac{1}{2}\cdot 1\right)\\
&= (-1)^1\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{1}\frac{(-1)^{k-1}}{k}\right)
\end{align*}
This proves the base case. If we assume that the identity is true for some $m\in\mathbb{N}$, then
\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan^{2(m+1)+1}(x)dx &= \int_{0}^{\frac{\pi}{4}}\tan^{2m+1+2}(x)\text{ }dx\\
&= \int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\tan^{2}(x)\text{ }dx\\
&= \int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\left[\sec^2(x)-1\right]dx\\
&= \int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\sec^2(x)\text{ }dx-\int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\text{ }dx
\end{align*}
Like before we make the substitution $u=\tan(x)$ in the first integral. Leveraging our assumption, we get
\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\sec^2(x)\text{ }dx-\int_{0}^{\frac{\pi}{4}}\tan^{2m+1}(x)\text{ }dx &= \int_{0}^{1}u^{2m+1}\text{ }du-(-1)^m\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\
&= \frac{1}{2m+2}-(-1)^m\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\
&= \frac{(-1)^m(-1)^m}{2(m+1)}-(-1)^m\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\
&= (-1)^m\left[\frac{(-1)^m}{2(m+1)}-\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\right]\\
&= (-1)^m\cdot -\left(-\frac{(-1)^{m}}{2(m+1)}+\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\
&= (-1)^{m+1}\left(\frac{1}{2}\ln(2)-\frac{(-1)^{(m+1)-1}}{2(m+1)}-\frac{1}{2}\sum_{k=1}^{m}\frac{(-1)^{k-1}}{k}\right)\\
&= (-1)^{m+1}\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{m+1}\frac{(-1)^{k-1}}{k}\right)
\end{align*}
This shows that it must be true for $m+1$. Thus, mathematical induction implies that the identity is true for all natural numbers.
We are now in a position to prove that $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$. Notice that
\begin{align*}
\left|\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right| &= \left|(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)\right|\\
&= \left|\int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx\right|\\
&= \int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx
\end{align*}
since $\tan^{2n+1}(x)\geq 0$ for $0\leq x <\pi/2$. We can bound the quantity
$$\left|\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right|$$
by making the substitution $x=\tan^{-1}(t)$ on the last integral and proceeding as follows
\begin{align*}
\int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx &= \int_{0}^{1}\frac{t^{2n+1}}{1+t^2}dt\\
&< \int_{0}^{1}t^{2n+1}\text{ }dt\\
&= \frac{1}{2n+2}\\
&= \frac{1}{2(n+1)}
\end{align*}
This gives the upper bound $\left|\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right|<\frac{1}{2(n+1)}$, or
$$\left|\ln(2)-\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right|<\frac{1}{n+1}$$
Letting $n\to\infty$ and applying the squeeze theorem immediately gives the desired result.
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$$
Q.E.D.
If correct, this argument is absolutely beautiful! Not only does it prove the equality
$$\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}=\ln(2)$$
in a completely elementary fashion, it also establishes this absolutely astounding result:
$$\int_{0}^{\frac{\pi}{4}}\tan^{2n+1}(x)\text{ }dx=(-1)^n\left(\frac{1}{2}\ln(2)-\frac{1}{2}\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}\right)$$
In particular, combining this with $\int_{0}^{\pi/4}\tan^{2n}(x)\text{ }dx=(-1)^n\left(\frac{\pi}{4}-\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}\right)$, we now have a recipe for evaluating
$$\int_{0}^{\frac{\pi}{4}}\tan^{n}(x)\text{ }dx$$
for any natural number $n$. How awesome it that?!
Let me know what you think! If you have any suggestions for optimizing the proof or identify some logical errors, please don't hesitate to share them with me!

REPLY [2 votes]: Concerning the integral
$$I_n=\int \tan^n(x)\,dx$$
$$x=\tan ^{-1}\left(\sqrt{t}\right) \implies I_n=\frac 12 \int \frac{t^{\frac{n-1}{2}}}{ (t+1)}\,dt=\frac{t^{\frac{n+1}{2}} }{n+1}\,\, _2F_1\left(1,\frac{n+1}{2};\frac{n+3}{2};-t\right)$$ where appears the Gaussian hypergeometric function.
$$J_n=\int_0^{\frac \pi 4} \tan^n(x)\,dx=\frac{1}{4} \left(\psi \left(\frac{n+3}{4}\right)-\psi\left(\frac{n+1}{4}\right)\right)$$
Expanded as series
$$J_n=\frac 1{2n}\sum_{k=0}^\infty \frac{E_{2 k}}{ n^{2 k}}$$ where appear Euler numbers.