Is there a closed-form of $ \sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$

Question

Is there a closed-form of $$\sum_{n=0}^{\infty }\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}$$

Thanks for any help

Answer 1

Let we sketch an alternative technique. Since: $$\sum_{n\geq 0} x^{2n} = \frac{1}{1-x^2} = \frac{1}{2}\left(\frac{1}{1+x}+\frac{1}{1-x}\right),\tag{1}$$ we have: $$\sum_{n\geq 0} \frac{x^{2n+1}}{2n+1} = \frac{1}{2}\left(\log(1+x)-\log(1-x)\right)=\operatorname{arctanh} x,\tag{2} $$ and integrating again: $$\sum_{n\geq 0} \frac{x^{2n+2}}{(2n+1)(2n+2)} = x\operatorname{arctanh} x+\frac{1}{2}\log(1-x^2),\tag{3} $$ $$\sum_{n\geq 0} \frac{x^{2n+3}}{(2n+1)(2n+2)(2n+3)} =\frac{1}{2}\left( -x+(x^2+1)\operatorname{arctanh} x+x\log(1-x^2)\right),\tag{4} $$ and by setting $g(x)=\sum_{n\geq 0} \frac{x^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)} $ we have: $$g(x)=\frac{1}{12}\left( -5x^2+(2x^4+6x^2)\frac{\operatorname{arctanh} x}{x}+(3x^2+1)\log(1-x^2)\right),\tag{5} $$ so: $$\sum_{n\geq 0}\frac{x^{n+2}}{(2n+1)(2n+2)(2n+3)(2n+4)}=\frac{1}{12}\left(-5x+(2x^2+6x)\frac{\operatorname{arctanh}\sqrt{x}}{\sqrt{x}}+(3x+1)\log(1-x)\right)\tag{6}$$ and finally: $$\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}=\color{red}{\frac{5-\pi-2\log 2}{12}}.\tag{7}$$

Answer 2

we know that the Taylor series of arctan(x) is $$\tan^{-1}(x)=\frac{(-1)^nx^{2n+1}}{2n+1}$$ take a triple integral for both sides $$\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}\tan^{-1}(x)dxdxdx=\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}\frac{(-1)^nx^{2n+1}}{2n+1}dxdxdx $$ $$\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}\tan^{-1}(x)dxdxdx=\frac{(-1^n)}{(2n+1)(2n+2)(2n+3)(2n+4)} $$

$$\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}\tan^{-1}(x)dxdxdx=\frac{1}{12}(5x^2+(1-3x^2)\log(x^2+1)+2(x^2-3)\tan^{-1}(x))|^{x=1}$$

$$\int_{0}^{1}\int_{0}^{x}\int_{0}^{x}\tan^{-1}(x)dxdxdx=\color{red}{\frac{1}{12}(5-2\log(2)-\pi )}$$

Answer 3

Let $$f(x)=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)}.$$

Then taking the derivatives wrt $x$, the factors in the denominator disappear: $$f''''(x)=\sum_{n=0}^{\infty }(-x^2)^n=\frac1{1+x^2}.$$

Integrate four times from $0$ to $x$ (factors added for convenience). $$\begin{align}f'''(x)&=\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+1}}{2n+1}\\ &=\arctan x.\end{align}$$ $$\begin{align}2f''(x)&=2\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}\\ &=2x\arctan x-\log(x^2+1).\end{align}$$ $$\begin{align}2f'(x)&=2\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+3}}{(2n+1)(2n+2)(2n+3)}\\ &=(x^2-1)\arctan x-x\log(x^2+1)+x.\end{align}$$ $$\begin{align}12f(x)&=12\sum_{n=0}^{\infty }\frac{(-1)^nx^{2n+4}}{(2n+1)(2n+2)(2n+3)(2n+4)}\\ &=2(x^2-3)x\arctan x-(3x^2-1)\log(x^2+1)+5x^2.\end{align}$$

Then $$f(1)=\dfrac{-\pi-2\log2+5}{12}.$$

For the same "price", you also get the sums with $1$ to $3$ factors: $f'''(1)=\pi/4$, $f''(1)=\pi/4-\log2/2$, $f'(1)=-\log2/2+1/2$.

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For efficient evaluation of the integrals, we can proceed as follows using complex numbers.

Notice that $\dfrac1{x^2+1}$ is the imaginary part of $\dfrac1{x-i}$, and use the change of variable $t=x-i$. Integrate four times (by parts) between $-i$ and $t$: $$\frac1t,$$ $$\log t,$$ $$t(\log t-1),$$ $$\frac{t^2}4(2\log t-3),$$ $$\frac{t^3}{36}(6\log t-11).$$ Then take the imaginary parts of the differences of the antiderivative values between $1-i$ and $-i$: $$\Im\left(\log(1-i)-\log(-i)\right)=\frac\pi4,$$ $$\Im\left((1-i)\left(\log(1-i)-1\right)-(-i)\left(\log(-i)-1\right)\right)=-\frac\pi4-\frac{\log2}2,$$ $$\Im\left(\frac{(1-i)^2}4(2\log(1-i)-3)-\frac{(-i)^2}4(2\log(-i)-3)\right)=\frac32-\pi-\frac{\ln2}2$$

$$\Im\left(\frac{(1-i)^3}{36}\left(6\log(1-i)-11\right)-\frac{(-i)^3}{36}\left(6\log(-i)-11\right)\right)=\frac\pi{12}-\frac{\log2}6-\frac{11}{12}$$

Answer 4

To give another approach:

The sum can be calculated using the observation that $$\int_0^1\left(1-x^{\frac{1}{n}}\right)^kdx=\frac{1}{\binom{n+k}{n}}$$ which is related to the beta function. Using this we get: $$\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)(2n+2)(2n+3)(2n+4)}=\frac{1}{4!}\sum_{n=0}^{\infty}\frac{(-1)^n}{\binom{2n+4}{4}}=\frac{1}{4!}\sum_{n=0}^{\infty}(-1)^n\int_0^1\left(1-x^{\frac{1}{4}}\right)^{2n}dx=\frac{1}{4!}\int_0^1\sum_{n=0}^{\infty}(-1)^n\left(1-x^{\frac{1}{4}}\right)^{2n}dx=\frac{1}{4!}\int_0^1\frac{1}{1+\left(1-x^{\frac{1}{4}}\right)^2}dx=\frac{1}{4!}\int_0^1\frac{4x^3}{1+\left(1-x\right)^2}dx=\frac{1}{3!}\int_0^1\frac{(1-x)^3}{1+x^2}dx=\frac{1}{3!}\int_0^1\frac{1-3x+3x^2-x^3}{1+x^2}dx=\frac{1}{6}\left[\arctan(x)-\frac32\ln\left(1+x^2\right)+3x-3\arctan(x)-\frac{x^2}{2}+\frac{1}{2}\ln\left(1+x^2\right)\right]_0^1=\frac{1}{6}\left(-\frac{\pi}{2}-\ln(2)+\frac{5}{2}\right)$$