Find value of $ S = \sum_{n=0}^{\infty} \frac{1}{n!(n+2)} $

Question

I am stuck on this question. Could someone help me?

$$ \text{Find value of } S = \displaystyle\sum_{n=0}^{\infty} \cfrac{1}{n!(n+2)} $$

I am supposed to show that $ S = 1 $ in two ways:

1) Integrate the taylor series of $ xe^x $
2) Differentiate the taylor series of $ \frac{e^{x-1}}{x} $

For (1), I tried to using the fact that the taylor series of $ e^x = \displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!} $

Now, multiplying $ x $ into the taylor series gives: $$ xe^x = \displaystyle\sum_{n=0}^{\infty} \cfrac{x^{n+1}}{n!} $$

Integrating this yields the following:

$$ \begin{align} \int_0^x xe^x &= \displaystyle\int_0^x\sum_{n=0}^{\infty} \cfrac{x^{n+1}}{n!} \\ &= \displaystyle\sum_{n=0}^{\infty} \cfrac{x^{n+2}}{n!(n+2)} \end{align} $$

I am not sure how to cary on from here.

For (2), I am not sure how to find the taylor series for $ \frac{(e^x - 1)}{x} $

Is anyone able to assist me?

Answer 1

For (1), you need to be careful with using $x$ as your dummy variable: try

$$\int_0^t xe^x\ dx = \int_0^t \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}\ dx$$

$$ = \sum_{n=0}^{\infty} \int_0^t\frac{x^{n+1}}{n!}\ dx$$

$$ = \sum_{n=0}^{\infty} \frac{x^{n+2}}{n!(n+2)}\bigg|_0^t$$

$$ = \sum_{n=0}^{\infty} \frac{t^{n+2}}{n!(n+2)}$$

At $t=1$, this equals our desired sum. We then evaluate the integral

$$\int_0^1 xe^x\ dx$$

which can be done with integration by parts.

For (2), we can simply use that the Taylor Series of $e^x$ is

$$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$$

$$e^x-1 = \sum_{n=1}^{\infty} \frac{x^n}{n!}$$

$$\frac{e^x-1}{x} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!} = \sum_{n=0}^{\infty} \frac{x^n}{(n+1)!}$$

and then differentiate, and set $x=1$ as you did before.

Answer 2

Observe \begin{align} \int^1_0 xe^x\ dx = \sum^\infty_{n=0} \int^1_0\frac{x^{n+1}}{n!}\ dx = \sum^\infty_{n=0} \frac{1}{n!(n+2)} \end{align}

Answer 3

Another way $$S = \sum_{n=0}^{\infty} \cfrac{1}{n!(n+2)}\\S = \displaystyle\sum_{n=0}^{\infty} \cfrac{1}{n!(n+2)}\times \frac{n+1}{n+1}\\=\sum_{n=0}^{\infty} \cfrac{n+1}{(n+2)!}\\= \sum_{n=0}^{\infty} \cfrac{(n+2)-1}{(n+2)!}\\= \sum_{n=0}^{\infty} \cfrac{n+2}{(n+2)!}-\cfrac{1}{(n+2)!}\\= \sum_{n=0}^{\infty} \cfrac{1}{(n+1)!}-\cfrac{1}{(n+2)!}\\= \sum_{n=0}^{\infty} f(n)-f(n+1)\\= \frac{1}{(0+1)!}-\lim_{n \to \infty}\frac{1}{(n+2)!}\\=1-0 $$