A Little Question from Math Twitter

I normally don’t write up full solutions for the little math problems people post on Twitter (now $\mathbb{X}$). This one was fun though.

$$\begin{equation} = \sum_{n=1}^{\infty} \frac{1}{n\Big(n + 1)\Big(n + 1)!} \end{equation}$$

First, we should establish a few useful facts. Facts $\Big(1)$ and $\Big(2)$ are properties of the gamma function. Fact $\Big(3)$ uses the Maclaurin expansion of $e^x$ to calculate the sum of inverse factorials.

$$\begin{align} \Big(1) & \Gamma\Big(n) = \Big(n - 1)! \\ \Big(2) & n\Gamma\Big(n) = \Gamma\Big(n + 1) \\ \Big(3) & \sum\limits_{n=1}^{\infty} \frac{1}{n!} = e - 1 \qquad & ,\qquad \textit{notice:} \sum\limits_{n=0}^{\infty} \frac{x^n}{n!} = e^x \\ \end{align}$$

Wolphram Alpha: Partial Fraction Calc. We can quickly approximate a solution by ignoring the $1/\Gamma\Big(n)$ factor. In this case, $\Big(A)$ evaluates to the Basel Sum ($\pi^2/ 6$), $\Big(C)$ evaluates to $\pi^2/6 - 1$, and the difference in $\Big(B)$ and $\Big(D)$ is just $-2$. Overall, $\pi^2/3 - 3 \approx 0.2898$, not such a bad estimate!

Removing an $\Big(n + 1)$ and an $n$ from the $\Big(n + 1)!$ gives the following. We can now use partial fractions to split the denominator into more managable terms.

$$\begin{align} = & \sum_{n=1}^{\infty} \frac{1}{n^2\Big(n + 1)^2\Gamma\Big(n)} \\ = & \sum_{n=1}^{\infty} \frac{1}{\Gamma\Big(n)} \left( \underbrace{\frac{1}{n^2}}_{\text{A}} + \underbrace{\frac{2}{\Big(n + 1)}}_{\text{B}} + \underbrace{\frac{1}{\Big(n + 1)^2}}_{\text{C}} - \underbrace{\frac{2}{n}}_{\text{D}} \right) \end{align}$$

The $\pm \sum\frac{1}{nn!}$ terms in $\Big(A)$ and $\Big(C)$ can be calculated just as we calculate $\Big(B)$. See: The Exponential Integral.

Taking each term one at a time we can calulate $\Big(B)$ and $\Big(D)$ quite easily. However, Attempting to calulate $\Big(A)$ and $\Big(C)$ is a bit more convoluted.

$$\begin{align} A) \qquad \sum_{n=1}^{\infty} \frac{1}{\Gamma\Big(n)n^2} = \sum_{n=1}^{\infty} \frac{1}{nn!} \end{align}$$

$$\begin{align} & B) \qquad & \sum_{n=1}^{\infty} \frac{2}{\Gamma\Big(n)\Big(n + 1)} \\ & & = \sum_{n=1}^{\infty} \frac{2}{\Big(n - 1)!\Big(n + 1)} \\ & & = \sum_{m=0}^{\infty} \frac{2}{m!\Big(m + 2)} \; , \qquad \textit{m := n - 1} \\ & & = 2 \sum_{m=0}^{\infty} \frac{1}{m!} \int_0^1{x^{m + 1}} dx \\ & & = 2 \int_0^1 \sum_{m=0}^{\infty} \frac{x^{m + 1}}{m!} = 2 \int_0^1 xe^x dx \\ & & = 2 \\ \end{align}$$

Wolphram Alpha: Partial Fraction Calc for $\Big(C)$

$$\begin{align} & C) \qquad & \sum_{n=1}^{\infty} \frac{1}{\Gamma\Big(n)\Big(n + 1)^2} \\ & & = -\sum_{n=1}^{\infty} \frac{1}{nn!} + \sum_{n=1}^{\infty} \frac{1}{n!} \\ & & = -\sum_{n=1}^{\infty} \frac{1}{nn!} + \Big(e - 1) \end{align}$$

$$\begin{align} & D) \qquad & -2\sum_{n=1}^{\infty} \frac{1}{\Gamma\Big(n)n} = -2\sum_{n=1}^{\infty} \frac{1}{n!} \\ & & = -2\Big(e - 1) \end{align}$$

Noticing that the $\pm \sum\frac{1}{nn!}$ terms from $\Big(A)$ and $\Big(C)$ sum to zero, we collect the remaining terms without ever needing to evaluate those sums.

$$\begin{equation} 2 + \Big(e - 1) - 2\Big(e - 1) = 3 - e \approx 0.2817 \qquad \qquad \blacksquare \end{equation}$$