# Investigate the convergence of $\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}f\left(n+1\right)}{n!}$

Define $$f(n)=n+\frac 1 {n^2}$$. Consider the following sum- $$\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}f\left(n+1\right)}{n!}$$ Investigate whether this series converges. If it does, then find the value to which it converges.

I have calculated that the general term is $$T_n=\left(-1\right)^{n+1}\frac{(n+1)^3+1}{(n+1)^2\times n!}$$ although this does not help in proceeding anymore.

It doesn't look like a very difficult question, but I can't get a hold over it.

Edit: I forgot to mention that I have already figured out that the series converges using Ratio Test. What I want to know, is the point to which it converges.

Stirling's Approximation has been suggested in the comments although I don't see how to use it to get the convergence point.

• Are you aware of the growth rate of factorial? Commented May 26, 2023 at 14:09
• @MostafaAyaz Stirling? Yes, I am. Commented May 26, 2023 at 14:17
• @AdamRubinson I forgot to mention that I could figure out the convergence using ratio test, what I couldn't do is the where it converges part. Commented May 26, 2023 at 14:18
• @AdamRubinson yes, of course. I apologise for the mistake. Commented May 26, 2023 at 14:20
• Why the downvote though? Commented May 26, 2023 at 14:20

Since you have the convergence, to have the result, you can split the sum since both of the terms are converging.

You have :

$$S = \sum_{n=1}^\infty (-1)^{n+1}\frac{n}{n!} + \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n!n^2}.$$

The first term can be rewritten as :

$$\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{(n-1)!} = \sum_{n=0}^\infty (-1)^{n+2}\frac{1}{n!} = (-1)^2e^{-1} = e^{-1}.$$

The first term is an hypergeometric function. From the notation of https://en.wikipedia.org/wiki/Generalized_hypergeometric_function , the second term can be rewritten as (wolfromalpha) :

$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n!n^2} = -_3F_3(1,1,1,2,2,2;-1) \approx -0.89.$$

Finally,

$$S = e^{-1}- {_3}F_3(1,1,1,2,2,2;-1).$$

Perhaps you can get a more simple proof. Hope it helps.

• I think that in the expression $S=$ it shoyld be a plus instead of a minus Commented May 28, 2023 at 3:39
• I guess it's the best one can do (+1)! Commented Jun 5, 2023 at 5:36

You can do the ratio test, your serie is even absolutely convergent.

You have : $$|\frac{a_{k+1}}{a_k}| = \frac{1}{n+1}\frac{n+2+\frac{1}{(n+2)^2}}{n+1+\frac{1}{(n+1)^2}}.$$

You can see that the second fraction tends to $$1$$ whereas the first clearly tends to 0. The ratio tends then to 0 and the serie converges.

• I hate to break it to you, but as I have included in the question, I forgot to mention that I have already figured out that the series converges using Ratio Test. What I want to know, is the point to which it converges. Commented May 27, 2023 at 4:47