# Calculate the series $1/(k+2)k!$

Let $$S = \sum_{k=0} ^\infty \frac{1}{(k+2)k!}$$. I am trying to evaluate this sum. I tried using the Taylor series of $$e^x$$, which is similar, but I am not sure how to deal with the $$1/k+2$$ factor. Any advice will be welcome.

• $\frac{1}{(k+2)k!} = \frac{k+1}{(k+2)!}$ – dEmigOd Feb 3 '19 at 16:32

You have

$$xe^x = \sum_{k=0}^\infty \frac{x^{k+1}}{k!}$$

By integration $$\int_0^x te^t \ dt = \sum_{k=0}^\infty \frac{x^{k+2}}{(k+2)k!}$$

Plugging in $$x=1$$, you get

$$\sum_{k=0}^\infty \frac{1}{(k+2)k!} = \int_0^1 te^t \ dt=[te^t]_0^1- \int_0^1 e^t \ dt=1$$

So we start with $$\frac{1}{(k+2)k!} = \frac{k+1}{(k+2)!}$$, then $$\frac{k+1}{(k+2)!} = \frac{1}{(k+1)!} - \frac{1}{(k+2)!}$$.

Then $$S = \sum\limits_{k=0}^{\infty} \frac{1}{(k+2)k!} = \sum\limits_{k=0}^{\infty} \left(\frac{1}{(k+1)!} - \frac{1}{(k+2)!}\right)$$

and this is telescopic.

Can you take it from here?

• Thanks. It helped, wish I could accept more than one answer – R. Davis Feb 3 '19 at 18:10

Have a look at the partial sums:

$$\sum_{k=0}^0 = \frac{1}{2} = \frac{2!-1}{2!}$$

$$\sum_{k=0}^1 = \frac{5}{6} = \frac{3!-1}{3!}$$

$$\sum_{k=0}^2 = \frac{23}{24} = \frac{4!-1}{4!}$$

$$\sum_{k=0}^3 = \frac{119}{120} = \frac{5!-1}{5!}$$

Here, you can detect a pattern and conclude $$\sum_{k=0}^n = \frac{(n+2)!-1}{(n+2)!} = 1 - \frac{1}{(n+2)!}$$. Hence you can calculate $$\lim\limits_{n \to \infty}\sum_{k=0}^n = \lim\limits_{n \to \infty} \left(1 - \frac{1}{(n+2)!}\right) = 1$$.