Series with arctan

Let $x \in (-1, 1)$. Evaluate the sum:

$$\sum_{n=1}^{\infty}(-1)^{n+1}n \left ( \tan^{-1}x - x + \frac{x^3}{3}-\cdots + (-1)^{n+1}\frac{x^{2n+1}}{2n+1}  \right )$$

(Proposed by Anastasios Kotronis at MathProblems)

Solution



We know that:

$$\arctan x =\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k-1}}{2k-1}= x - \frac{x^3}{3}+ \frac{x^5}{5}+\cdots $$

Hence:

$$\arctan x- \sum_{k=1}^{n} \frac{(-1)^{n+1}x^{2k-1}}{2k-1}= \sum_{k=n+1}^{\infty}\frac{(-1)^{n+1}x^{2k-1}}{2k-1}$$

Therefore the initial series can be expressed as:

$$\sum_{n=1}^{\infty}(-1)^{n+1}n \sum_{k=n+1}^{\infty}\frac{(-1)^{k+1}x^{2k-1}}{2k-1}$$

However, the following hold:

$$\begin{aligned}\sum_{n=1}^{\infty}a_n \sum_{k=n+1}^{\infty}b_k &=a_1 \left ( b_2 + b_3 +\cdots \right ) +a_2 (b_3 +b_4 +\cdots)\\
&=b_2 a_1 + b_3 \left ( a_1 +a_2 \right )+ \cdots \\
&=\sum_{k=2}^{\infty}b_k \sum_{n=1}^{k-1}a_n \\ 
&= \sum_{k=2}^{\infty}b_k\left ( \sum_{n=1}^{k}a_n - a_k \right )\\
&= \sum_{k=2}^{\infty} b_k \sum_{n=1}^{k}a_n - \sum_{k=2}^{\infty}b_k a_k\\
&= \sum_{k=1}^{\infty}b_k \sum_{n=1}^{k}a_n - \sum_{k=1}^{\infty}b_k a_k 
\end{aligned}$$

Then:

$$\begin{aligned}
\sum_{n=1}^{\infty}(-1)^{n+1}n \sum_{k=n+1}^{\infty}\frac{(-1)^{n+1}x^{2k-1}}{2k-1} &=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k-1}}{2k-1}\sum_{n=1}^{k}(-1)^{n+1}n - \\
& \quad \quad \quad \quad \quad -\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k-1}}{2k-1}(-1)^{k+1} \\
&=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k-1}}{2k-1}\sum_{n=1}^{k}(-1)^{n+1}n - \\
& \quad \quad \quad \quad \quad - \sum_{k=1}^{\infty}\frac{k x^{2k-1}}{2k-1} \\
\end{aligned}$$

However,

$$- \sum_{k=1}^{\infty}\frac{k x^{2k-1}}{2k-1}= - \frac{1}{2}\sum_{k=1}^{\infty}\left ( 1 + \frac{1}{2k-1} \right )x^{2k-1}= - \frac{x}{2(1-x^2)}+ \frac{1}{4}\log \frac{1-x}{1+x}$$

and for the other sum by splitting up to even and odd terms we have that:

$$\begin{aligned}
\sum_{k=1}^{\infty}\frac{(-1)^{k+1}x^{2k-1}}{2k-1}\sum_{n=1}^{k}(-1)^{n+1}n &=\sum_{k=1}^{\infty}\frac{(-1)^{2k}x^{4k-1}}{4k-1}\sum_{n=1}^{2k}(-1)^n n + \\
& \quad \quad \quad \quad + \sum_{k=1}^{\infty}\frac{(-1)^{2k-1}x^{4k-3}}{4k-3}\sum_{n=1}^{2k-1}(-1)^n n \\
&= \sum_{k=1}^{\infty}\frac{k x^{4k-1}}{4k-1} + \sum_{k=1}^{\infty}\frac{k x^{4k-3}}{4k-3}\\ &= \sum_{k=1}^{\infty}k \int_{0}^{x}y ^{4k-2}\, {\rm d}y + \sum_{k=1}^{\infty}k \int_{0}^{x}y^{4k-4}\, {\rm d}y \\
&\overset{{\rm monotone \; convergence \; theorem}}{=\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! =\! }\int_{0}^{x}\sum_{k=1}^{\infty}k y^{4k-2}\, {\rm d}y + \\
& \quad \quad \qquad \qquad + \int_{0}^{x} \sum_{k=1}^{\infty}k y^{4k-4}\, {\rm d}y \\&= \int_{0}^{x}\frac{y^2}{\left ( 1-y^4 \right )^2}\, {\rm d}y + \int_{0}^{x}\frac{{\rm d}y}{\left ( 1-y^4 \right )^2} \\
&= \int_{0}^{x}\frac{1+y^2}{\left ( 1-y^4 \right )^2}\, {\rm d}y \\
&= \frac{1}{4}\left ( \log \frac{1-x}{1+x}+ \arctan x + \frac{x}{1-x^2} \right )
\end{aligned}$$

Hence the original sum evaluates to:

$$\sum_{n=1}^{\infty}(-1)^{n+1}n \left ( \arctan x - x +\frac{x^3}{3} - \cdots + (-1)^{n+1}\frac{x^{2n+1}}{2n+1} \right )= \frac{1}{4}\arctan x - \frac{x}{4 \left ( 1-x^2 \right )}$$

and the exercise comes to and end.

The exercise can also be found in mathematica.gr