# Integral of a sum converges or diverges?

Liouville's constant is an artificially-constructed mathematical constant defined as:

$$\displaystyle\sum^\infty_{n=1}10^{-n!}=0.\mathbf{1}\mathbf{1}000\mathbf{1}00000000000000000\mathbf{1}00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000\mathbf{1}\ldots$$

Liouville's constant can be constructed in base $$b$$ by replacing the $$10^{-n!}$$ with $$b^{-n!}$$.

Now let's consider the sum that produces Liouville's Constant in base $$b$$: $$\displaystyle\sum_{n\in\mathbb{Z}^{+}}b^{-n!}$$

And finally, the question I was getting at. Is the following integral:

$$\displaystyle\int^{1}_{0}\left(\sum_{n\in\mathbb{Z}^{+}}x^{n!}\right)dx$$

convergent or divergent?

I honestly don't even know where to start. How would I go about estimating the value of this integral if it converges?

## 1 Answer

First interchange the order of the integration and the summation to get $$\sum_1^\infty\int_0^1 x^{n!}dx=\sum_1^\infty\frac{1}{n!+1}$$ Now, $$n!\lt n!+1$$ so $$\frac{1}{n!+1}\lt\frac{1}{n!}$$ this gives $$\sum_1^\infty\frac{1}{n!+1}\lt\sum_1^\infty\frac{1}{n!}=e-1$$ by the series definition of e, thus the series converges.