How to evaluate the following? $\sum\limits_{n=1}^{\infty} \int_1^{n!} x^{-n}{\rm d}x$.

Question

How can I evaluate this sum?

$$\sum_{n=1}^{\infty} \int_1^{n!} x^{-n} \mathrm{d} x$$ Is there a closed form or a transform that would make it possible not to "CAS" it? Also, it seems to converge, but I cannot prove it.

Answer 1

How can I evaluate this sum?

Start by evaluating the painstakingly simple definite integral inside the series.

Is there a closed form

Yes, $\infty$.

Also, it seems to converge

It sure does, doesn't it ? But you know what they say about appearances... ;-)

but I cannot prove it.

I'd be amazed if you could. :-) Well, if you can't prove that it converges, prove that it diverges. ;-)

Answer 2

I would like to think that the sum doesn't converge. Below is the intuitive proof; I think someone else can make this rigorous or prove this to be wrong.

Firstly, for $n=1$,$\int_1^1 x^{-1} dx = 0$. Next, for $n >1$,

$$I_n = \int_{1}^{n!}x^{-n}dx =\frac{1}{-n+1}x^{-n+1}|_1^{n!} = \frac{1}{1-n}[n!^{1-n}-1] = \frac{1}{n-1}\left[1-\frac{1}{n!^{n-1}} \right] $$

as $n\rightarrow\infty$, then $\left[1-\frac{1}{n!^{n-1}}\right]\rightarrow 1$ in a really quick way, since $n!^{n-1}$ explodes fastly; the relevant factor is that $\frac{1}{n!^{n-1}}$ converges to $0$ much, much faster compared to $\frac{1}{n}$ converging to 0.

Thus we can say that $(n-1) I(n)\rightarrow 1$ as $n\rightarrow\infty$, that is, the behaviour of $I(n)$ highly resembles $\frac{1}{n-1}$ as we move to larger values of $n$.

However, we know that $\sum_2^\infty\frac{1}{n-1}$ doesn't converge, which leads me to believe that the sum doesn't converge.

(Hope someone can fix my write-up; don't really now how to express this idea!)

Answer 3

Remove the $n=1$ term. The rest is more than $\sum_{n=2}^{\infty}\int_1^2x^{-n}dx$