# Comparison Test

• Nov 13th 2009, 06:17 AM
CoraGB
Comparison Test
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!

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).
• Nov 13th 2009, 08:10 AM
tonio
Quote:

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).

Even better, perhaps: you can choose $\displaystyle \forall\,\epsilon>0\,\,N_{\epsilon}\in\mathbb{N}\, \,s.t.\,\,\forall\,n>N_{\epsilon}\,,\,\,\sum\limit s_{n=N_{\epsilon}+1}^\infty\frac{1}{n!}<\epsilon$, and then:

$\displaystyle \sum\limits_{n=N_{\epsilon}+1}^\infty\frac{\{n!x\} }{n!}-\sum\limits_{n=N_{\epsilon}+1}^\infty\frac{\{n!a\} }{n!}=$ $\displaystyle \sum\limits_{n=N_{\epsilon}+1}^\infty\frac{1}{n!}\ left(\{n!x\}-\{n!a\}\right)\leq\sum\limits_{n=N_{\epsilon}+1}^\ infty\frac{1}{n!}<\epsilon$

Tonio
• Nov 15th 2009, 05:51 AM
CoraGB
Thank you so much, especially with part b.

I wwas wondering if you could then help me with showing that f(x) is not differentiable at x=0.

All I know so far is that with a sequence a_n goes to zero,
[g(a_n) – g(0)]/[a_n – 0] = g(a_n)/a_n≥ 1
• Nov 15th 2009, 06:54 AM
tonio
Quote:

Originally Posted by CoraGB
Thank you so much, especially with part b.

I wwas wondering if you could then help me with showing that f(x) is not differentiable at x=0.

All I know so far is that with a sequence a_n goes to zero,
[g(a_n) – g(0)]/[a_n – 0] = g(a_n)/a_n≥ 1

If the function's differentiable at zero then we have:

$\displaystyle f'(0):=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\sum\limits_{n=0}^\infty\frac{\{n!x\}}{xn!}$ , and this limit must exist no matter how we choose to make $\displaystyle x\rightarrow 0$, so:

1) Choose to evaluate the limit through the sequence $\displaystyle \left\{\frac{1}{k}\right\}_{k=1}^\infty$, and check what happens with the sum when $\displaystyle n\geq k$ (you can do this since convergence/divergence doesn't change when we throw away a finite number of summands in a series)

2) Now choose to evaluate the limit through the sequence $\displaystyle \left\{\frac{\pi}{k}\right\}_{k=1}^\infty$...

Tonio