Calculate $\sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+3)}\cdot\frac{1}{n + 1}$

Question

I’m trying to calculate the following sum $$ \sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+3)}\cdot\frac{1}{n + 1} $$

I know that $$ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} = \ln(2)\\ \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{\pi}{4}\\ \sum_{n=0}^{\infty} \frac{1}{(4n+1)(4n+3)} = \frac{1}{2}\cdot\sum_{n=0}^{\infty} \left(\frac{1}{4n+1} - \frac{1}{4n+3}\right) = \frac{1}{2}\left(1 - \frac{1}{3} +\frac{1}{5} -\dots\right) = \frac{\pi}{8}\\ \sum_{n=0}^{\infty}\frac{1}{(4n+2)(4n+4)} = \frac{1}{2}\left(\frac{1}{2} - \frac{1}{4} + \frac{1}{6} -\dots\right) = \frac{1}{4} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n+1} = \frac{\ln(2)}{4} $$

Could these calculations lead to the answer? And if not what is a better approach?

Answer 1

$$S=\sum_{n=0}^{\infty} \frac1{(4n+1)(4n+3)(n+1)} = \frac13\sum_{n=0}^{\infty} \left( \frac2{4n+1}-\frac6{4n+3}+\frac4{4n+4}\right)$$

We have that $$\sum\left(\frac1{4n+1}-\frac1{4n+2}+\frac1{4n+3}-\frac1{4n+4}\right)=\ln2$$$$\sum\left(\frac4{4n+1}-\frac4{4n+3}\right)=\pi$$

$$\begin{align} 3S+3\ln2&=\pi+\sum_{n=0}^{\infty} \left(\frac1{4n+1}-\frac3{4n+2}+\frac1{4n+3}+\frac1{4n+4}\right)\\ &=\pi+\sum_{n=0}^{\infty} \left(\left(\frac1{4n+1}-\frac1{4n+2}+\frac1{4n+3}-\frac1{4n+4}\right)-\left(\frac2{4n+2}-\frac2{4n+4}\right)\right) \end{align}$$

Since both brackets are convergent we can separate the summation

$$ \begin{align} 3S+3\ln2&=\pi+\sum_{n=0}^{\infty} \left(\frac1{4n+1}-\frac1{4n+2}+\frac1{4n+3}-\frac1{4n+4}\right)-\sum_{n=0}^{\infty}\left(\frac1{2n+1}-\frac1{2n+2}\right)\\ &=\pi+\ln2-\ln2=\pi \end{align} $$

$$ S=\frac13\left(\pi-3\ln2\right)\approx 0.354050371 $$

Answer 2

It reminds me of the digamma function. So, I'll try a non-trivial approach to calculate the given infinite sum. I hope you'll find it interesting.

We have the series formula for the digamma function. $$\psi(z)=-\gamma+\sum_{n=0}^\infty\frac{z-1}{(n+1)(n+z)},\qquad z\notin\mathbb{Z}_{\leq0}\tag{1}$$

Substitute $z=\frac14$: $$\psi\left(\frac14\right)=-\gamma-\sum_{n=0}^\infty\frac{3}{(n+1)(4n+1)}\implies\color{red}{\sum_{n=0}^\infty\frac{1}{(n+1)(4n+1)}=-\frac{\gamma+\psi(1/4)}{3}}\tag{2}$$ Similarly, $z=\frac34$:

$$\color{red}{\sum_{n=0}^\infty\frac{1}{(n+1)(4n+3)}=-\gamma-\psi(3/4)} \tag{3}$$ We know the digamma of $1/4$ from the digamma additive formula or Gauss' digamma theorem. $$\psi\left(\frac14\right)=-\gamma - 3 \ln 2 - \dfrac \pi 2\tag{4}$$ Reflection formula for the digamma function: $$\psi(1-z)-\psi(z)=\pi\cot\pi z\implies\psi\left(\frac34\right)-\psi\left(\frac14\right)=\pi\tag{5}$$

Now, $$\begin{aligned}\sum_{n=0}^{\infty} \frac1{(4n+1)(4n+3)(n+1)}&=\frac12\cdot\sum_{n=0}^\infty\left(\frac{1}{(n+1)(4n+1)}-\frac{1}{(n+1)(4n+3)}\right)\\&=\frac{-\gamma-\psi(1/4)+3\gamma+3\psi(3/4)}{6}\\&=\frac{2\gamma+3\pi-2\gamma-6\ln2-\pi}{6}\\\color{red}{\sum_{n=0}^{\infty} \frac1{(4n+1)(4n+3)(n+1)}}&\color{red}{=\frac\pi3-\ln2\approx0.35405} \end{aligned}$$

---

$$\begin{aligned}\sum_{n=0}^{\infty} \frac1{(4n+1)(4n+3)(n+1)}&=\sum_{n=0}^\infty\left(\frac{2}{3(4n+1)}-\frac{2}{4n+3}+\frac{1}{3(n+1)}\right)\\&=\frac43\sum_{n=0}^\infty\left(\frac{1}{4n+1}-\frac{1}{4n+3}\right)-\frac23\sum_{n=0}^\infty\left(\frac{1}{2(2n)+1}+\frac{1}{2(2n+1)+1}-\frac{1}{2(n+1)}\right)\\&=\frac43\cdot\frac\pi4-\frac23\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)-\frac23\sum_{n=0}^\infty\left(\frac{1}{2(2n)+2}+\frac{1}{2(2n+1)+2}-\frac{1}{2n+2}\right)\\&=\frac\pi3-\frac23\cdot\ln2-\frac13\sum_{n=0}^\infty\left(\frac{1}{2n+1}-\frac{1}{2n+2}\right)\\&=\frac\pi3-\frac23\cdot\ln2-\frac13\ln2\\&=\frac\pi3-\ln2 \end{aligned}$$

Answer 3

Building on your approach with partial fraction decomposition, you can do the following: \begin{align} \frac{1}{(4𝑛+1)(4𝑛+3)}\cdot \frac{1}{n+1} &= \frac{1}{2}\left(\frac{1}{4𝑛+1}-\frac{1}{4𝑛+3}\right)\cdot \frac{1}{n+1} \\ &= \frac{1}{2}\left(-\frac{1}{3}\left(\frac{1}{n+1}-\frac{4}{4n+1}\right)+\left(\frac{1}{n+1}-\frac{4}{4n+3}\right)\right) \end{align}

From here you can sum from 0 to $\infty$ and get some geometric series sums that work more in your favor.