Originally Posted by

**CoraGB** I was wondering if anyone could help me with the following problems.

I am working with the function f(x) = Σ ({n!*x})/n! where n goes from 0 to infintity and {x} = distance from x to the nearest integer.

a) I need to use the comparison test to show that f(x) converges. I know that I can compare it to Σ 1/n!

Exactly. The function $\displaystyle \{x\}\,,\,\,x\in\mathbb{R}$ is also called the fractional part of x function, and it is true that $\displaystyle \forall\,x\in\mathbb{R}\,,\,\,0\leq \{x\}\leq 1$, so $\displaystyle \frac{\{n!x\}}{n!}\leq \frac{1}{n!}$ and since the series of the right hand sequence clearly converges (and everything's non-negative here) we get that our series converges.

b) I need to show that f(x) is continuous at x=a. I know that I can chose an N so that Σ ({n!*x})/n! < ε (n goes from N+1 to infinity), and then use

Σ ({n!*x})/n! - Σ ({n!*a})/n! (n goes from N to infinity).