# Thread: Convergence of Series of Functions

1. ## Convergence of Series of Functions

Hello!

I am working on what seems to be a very "concrete" exercise in showing convergence.

on the interval $[0,1]$, show that the functions converge to the function f(x) point-wise, and determine if they converge uniformly.

a) $\{f_n(x)\}_{n=1}^{\infty} = \{\frac{2x}{1+nx}\}$ where $f(x) = 0$

b) $\{f_n(x)\}_{n=1}^{\infty} = \{\frac{cos(nx)}{\sqrt{n}}\}$ where $f(x) = 0$

c) $\{f_n(x)\}_{n=1}^{\infty} = \{\frac{n^3x}{1+n^4x}\}$ where $f(x) = 0$

What I've got so far...

So, to show a function is convergent point-wise, we just try to show that the NUMBER sequence $\{f_n(x)\}$ converges to $f(x) = 0$ so, $\forall \epsilon > 0, \exists N$ such that $n > N$ implies $f_n(x) - 0 < \epsilon$

Here is where I get confused. Do we define an epsilon in terms of n?! Which would depend on the series of functions in all 3 cases?
And for uniform convergence, we are seeking a similar result but with an epsilon which is independent of n?

Any help/guidance to set up the question further appreciated!!

2. I assume this are sequences of functions as opposed to series. In that case, regarding your questions, for pointwise convergence we just need an N dependent on x and n. For uniform convergence we need an N dependent only on n (The epsilon is arbitrary so it makes no sense that it would depend on something). For example in the second one we have the bound $0\leq \frac{|\cos(nx)|}{\sqrt{n}} \leq \frac{1}{\sqrt{n}}$ from which one sees that one can make the sequence of functions arbitrarily arbtrarily close to 0 independently of any point.

3. Yes... of course! Sorry, sequences of functions, not series. Thank you for pointing that out.

Also, of course, I was trying to find epsilon as a function of N but that's ridiculous, I'm not sure how my mind slipped on such basic epsilon-delta arguments. I'll blame it on an algebra OD.

Thanks!