Convincing that $\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n-2}(2n+1)(n!)} \ne \pi$

Question

A friend of mine in high school challenged me to calculate the value of the sum

$$\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n-2}(2n+1)(n!)}$$ And then claimed that the answer was $\pi$ . But when I worked out the sum myself I found the closed form $\\ 2 \sqrt{2 \pi}\ \text{Erf}(\frac{1}{\sqrt2})$. That guy knows nothing about the error function nor lot about manipulation of series . Is there any easy way to show him that that sum is not equal to $\pi$. ?

Answer 1

If we denote $a_n$ to be equal to

$$4 \sum_{k=0}^n \frac{(-1)^k}{2^k (2k+1) \cdot k!},$$

then $a_1 = 10/3$ and $a_2 = 103/30$, additionaly we have $a_1 \le a_n \le a_2$ for all $n$ (because that's an alternating series). Clearly $\pi < a_1$.

Answer 2

Your series converges absolutely so you can rearrange in this way $$4\sum_{n\geq0}\frac{\left(-1\right)^{n}}{2^{n}\left(2n+1\right)n!}=\left(4-\frac{2}{3}\right)+\left(\frac{1}{10}-\frac{1}{168}\right)+\dots$$ so each of the therm inside the parenthesis is positive. Now note that $$4-\frac{2}{3}=\frac{10}{3}=3.333\dots>\pi.$$