The series converges to?

Question

Consider the series $\sum a_n$ with $n$-th term:

$$a_{n}=\frac{1+3+\cdots+(2n-1)}{n!}.$$

It is required to obtain the convergence of the series. I am not able to start. I only know that this series will be increasing and is bounded below by $0$.

Answer 1

Since $$1+3+\cdots +2n-1 =n^2$$

Your $a_n$ can be re-written as $$a_n = \frac{n^2}{n!}= \frac{n \color{blue}{-1+1}}{(n-1)!}= \frac{1}{(n-2)!}+\frac1{(n-1)!}$$

Therefore, $$\sum_{n=1}^\infty a_n = \sum_{n=2}^\infty\frac{1}{(n-2)!}+\sum _{n=1}^\infty\frac 1{(n-1)!} =e+e=2e$$

Answer 2

Using arithmetic progression we have $$a_n=\frac{n^2}{n!}$$

Let $$A(x)=\sum_{n=0}^{\infty}\frac{1}{n!}x^n=e^x$$ Differentiate both sides we have $$A'(x)=\sum_{n=1}^{\infty}\frac{n}{n!}x^{n-1}=e^x\implies xA'(x)=\sum_{n=1}^{\infty}\frac{n}{n!}x^n=xe^x$$ Differentiate both sides again we have $$A''(x)=\sum_{n=1}^{\infty}\frac{n^2}{n!}x^{n-1}=\sum_{1=0}^{\infty}a_nx^{n-1}=(x+1)e^x$$ The sum converges for all $x\in\mathbb{R}$, and for $x=1$ we get $$\sum_{n=1}^{\infty}a_n=2e$$