Find the closed form for this series

Question

I found this interesting series from the , it is from an old math books. It is as followed:

$\dfrac{1}{2}-\dfrac{x^2}{6}+\dfrac{x^4}{12}-\dfrac{x^6}{20}+\dfrac{x^8}{30}-...$

I notice that one can rewrite this series as followed:

$\dfrac{1}{2}-\dfrac{x^2}{2\cdot 3}+\dfrac{x^4}{3\cdot 4}-\dfrac{x^6}{4\cdot 5}+\dfrac{x^8}{5\cdot 6}-...$

So the general formula for this series is

$$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}$$

Is there a closed form for this series?

Answer 1

Start with a geometric series \begin{eqnarray*} \sum_{n=0}^{\infty} (-1)^n y^n = \frac{1}{1+y}. \end{eqnarray*} Integerate \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{(-1)^n y^{n+1}}{n+1} = \ln(1+y). \end{eqnarray*} Integerate again \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{(-1)^n y^{n+2}}{(n+1)(n+2)} = (1+y)\ln(1+y)-y. \end{eqnarray*} Now divide by $y^2$ and let $y=x^2$ and we have \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(n+1)(n+2)} = \frac{(1+x^2)\ln(1+x^2)-x^2}{x^4}. \end{eqnarray*}

Answer 2

I get $(x^{-2}+x^{-4})\ln(1+x^2)-x^{-2} $.

I use $\ln(1+x) =\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^n}{n} $.

$\begin{array}\\ f(x) &=\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}\\ &=\sum_{n=0}^{\infty} (-1)^{n}x^{2n}(\dfrac1{n+1}-\dfrac1{n+2})\\ &=\sum_{n=0}^{\infty} (-1)^{n}x^{2n}\dfrac1{n+1}-\sum_{n=0}^{\infty} (-1)^{n}x^{2n}\dfrac1{n+2}\\ &=\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n-2}}{n}-\sum_{n=2}^{\infty} \dfrac{(-1)^{n+2}x^{2n-4}}{n}\\ &=x^{-2}\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}-x^{-4}\sum_{n=2}^{\infty} \dfrac{(-1)^{n}x^{2n}}{n}\\ &=x^{-2}\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}+x^{-4}\sum_{n=2}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n}\\ &=x^{-2}\ln(1+x^2)+x^{-4}(-x^2+\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^{2n}}{n})\\ &=x^{-2}\ln(1+x^2)+x^{-4}(-x^2+\ln(1+x^2))\\ &=(x^{-2}+x^{-4})\ln(1+x^2)-x^{-2}\\ \end{array} $

Answer 3

And here's my attempt to generalize this as much as possible.

Let $f(x) =\sum_{n=0}^{\infty} (-1)^{n}\sum_{k=1}^m \dfrac{a_kx^{c_kn}}{n+b_k} $ where the $b_k$ are positive integers and the $a_k$ and $c_k$ are real.

Then

$\begin{array}\\ f(x) &=\sum_{n=0}^{\infty} (-1)^{n}\sum_{k=1}^m \dfrac{a_kx^{c_kn}}{n+b_k} \\ &=\sum_{k=1}^m \sum_{n=0}^{\infty} (-1)^{n}\dfrac{a_kx^{c_kn}}{n+b_k} \\ &=\sum_{k=1}^m a_k\sum_{n=b_k}^{\infty} (-1)^{n-b_k}\dfrac{x^{c_k(n-b_k)}}{n} \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}\sum_{n=b_k}^{\infty} (-1)^{n-b_k}\dfrac{x^{c_kn}}{n} \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\sum_{n=b_k}^{\infty} (-1)^{n+1}\dfrac{x^{c_kn}}{n} \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}(\sum_{n=1}^{\infty} (-1)^{n+1}\dfrac{x^{c_kn}}{n}-\sum_{n=1}^{b_k-1} (-1)^{n+1}\dfrac{x^{c_kn}}{n}) \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}(\ln(1+x^{c_k})-\sum_{n=1}^{b_k-1} (-1)^{n+1}\dfrac{x^{c_kn}}{n}) \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\ln(1+x^{c_k})-\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\sum_{n=1}^{b_k-1} (-1)^{n+1}\dfrac{x^{c_kn}}{n} \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\ln(1+x^{c_k})-\sum_{k=1}^m a_k\sum_{n=1}^{b_k-1} (-1)^{n+b_k}\dfrac{x^{c_k(n-b_k)}}{n} \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}(-1)^{b_k+1}\ln(1+x^{c_k})-\sum_{k=1}^m a_k\sum_{n=1}^{b_k-1} (-1)^{n}\dfrac{x^{-c_kn}}{b_k-n} \\ \end{array} $

Similarly, let $f(x) =\sum_{n=0}^{\infty} \sum_{k=1}^m \dfrac{a_kx^{c_kn}}{n+b_k} $ where the $b_k$ are positive integers and the $a_k$ and $c_k$ are real.

Here I use $-\ln(1-x) =\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}x^n}{n} $.

Then

$\begin{array}\\ f(x) &=\sum_{n=0}^{\infty} \sum_{k=1}^m \dfrac{a_kx^{c_kn}}{n+b_k} \\ &=\sum_{k=1}^m \sum_{n=0}^{\infty}\dfrac{a_kx^{c_kn}}{n+b_k} \\ &=\sum_{k=1}^m a_k\sum_{n=b_k}^{\infty} \dfrac{x^{c_k(n-b_k)}}{n} \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}\sum_{n=b_k}^{\infty} \dfrac{x^{c_kn}}{n} \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}\sum_{n=b_k}^{\infty} \dfrac{x^{c_kn}}{n} \\ &=\sum_{k=1}^m a_kx^{-b_kc_k}(\sum_{n=1}^{\infty} \dfrac{x^{c_kn}}{n}-\sum_{n=1}^{b_k-1} \dfrac{x^{c_kn}}{n}) \\ &=-\sum_{k=1}^m a_kx^{-b_kc_k}(\ln(1-x^{c_k})-\sum_{n=1}^{b_k-1} \dfrac{x^{c_kn}}{n}) \\ &=-\sum_{k=1}^m a_kx^{-b_kc_k}\ln(1-x^{c_k})-\sum_{k=1}^m a_kx^{-b_kc_k}\sum_{n=1}^{b_k-1} \dfrac{x^{c_kn}}{n} \\ &=-\sum_{k=1}^m a_kx^{-b_kc_k}\ln(1-x^{c_k})-\sum_{k=1}^m a_k\sum_{n=1}^{b_k-1} \dfrac{x^{c_k(n-b_k)}}{n} \\ &=-\sum_{k=1}^m a_kx^{-b_kc_k}\ln(1-x^{c_k})-\sum_{k=1}^m a_k\sum_{n=1}^{b_k-1} \dfrac{x^{-c_kn}}{b_k-n} \\ \end{array} $

Answer 4

$$\sum_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2n}}{(n+1)(n+2)}=\sum_{n=0}^\infty(-1)^nx^{2n}\int_0^1\int_0^1 y^nz^{n+1}dydz$$

$$=\int_0^1\int_0^1z\sum_{n=0}^\infty(-x^2yz)^ndydz$$

$$=\int_0^1\int_0^1\frac{z}{1+x^2yz}dydz$$

$$=\int_0^1z\left(\int_0^1\frac{dy}{1+x^2yz}\right)dz$$

$$=\int_0^1z\left(\frac{\ln(1+x^2z}{x^2z}\right)dz$$

$$=\frac1{x^2}\int_0^1\ln(1+x^2z)dz$$

$$=\frac1{x^2}\cdot\frac{(1+x^2)\ln(1+x^2)-x^2}{x^2}$$