# Difference between sequence of functions and series

I am currently studying uniform and pointwise convergence and I am stuck at a somewhat basic distinction. Until now our lecturer has been talking about sequences of functions. For example:

$$f_n(x)=\frac{n}{x} \space \space \text{or} \space \space f_n(x)=x^n$$

However, in our problem set I am supposed to test the following series for uniform convergence:

$$\sum_{n=0}^{\infty}\frac{\sin(nt)}{e^n}$$

Are sequences of functions and series related? Is the process of showing uniform convergence different for series?

From any series $\sum_{k=0}^{\infty} a_k$, you can construct the sequence of partial sums, $$S_n = \sum_{k=0}^n a_k,$$ and treat it as you would a sequence. Conversely, given any sequence $(b_n)_{n=0}^{\infty}$, the differences $$a_0 = b_0, \quad a_k = b_k-b_{k-1}$$ can be made into a series, $$b_n = \sum_{k=0}^{n} a_k.$$ This is much like differentiation and integration being "inverse operations" (scare quotes required for technical reasons, of course).

• So in your example $S_1$ would be equal to $\sum_{k=0}^{1}a_K$; $S_2=\sum_{k=0}^{2}a_K$? Am I basically looking at the sequence of sums? – qmd Jun 3 '15 at 18:39
• Aside: What is a series itself? I think that a sequence is well-defined mathematical object but series is not. – user87690 Jun 3 '15 at 18:42
• Why is a series not well-defined? – qmd Jun 3 '15 at 18:47
• @Rzeta: So how do you define a series? – user87690 Jun 3 '15 at 18:52
• $\sum_0^n a_n=a_0+a_1+a_2+a_3+...+a_n$? – qmd Jun 3 '15 at 18:54

You can view a series as a sequence of partial sums. Try to think of the sequence of functions $$f_i(t)=\sum_{n=0}^i \frac{\sin(nt)}{e^n}$$

• How would I prove/disprove uniform convergence in this case. As far as I know I always need some sort of a limit function when I am trying to test uniform convergence but In this case I am not sure how to even get to that limiting function. – qmd Jun 3 '15 at 18:53
• @Rzeta: You don't need the limiting function. There are theorems you can use to prove that a series is convergent. In this case we have $|\sin(nt) / e^n| ≤ 1 / e^n$ and $∑_n 1 / e^n$ converges, so the original series converges uniformly. – user87690 Jun 3 '15 at 19:12
• @user87690 Can you name me some of the theorems. Is the Weierstrass-M-test one of them? I just came accross that while researching. Also, is there any way I can apply things like the ratio test or the root test when testing for uniform convergence? – qmd Jun 3 '15 at 19:21
• @Rzeta: Yes, Weierstrass-M-test is one of them and that's actually what I've used above. It says that if $|f_n(t)| ≤ a_n$ for every $n, t$ and $∑_n a_n$ converges, then $∑_n f_n(t)$ converges uniformly. You use things like the ratio test and the root test to show that your series $∑_n a_n$ chosen for Weierstrass actually converges. – user87690 Jun 3 '15 at 19:29
• That actually helped me a ton. Things are starting to fall into place ;). So basically I just have to be "skilled" at choosing an appropriate convergent $\sum_n a_n$ for the Weierstrass-M-test and I am done? – qmd Jun 3 '15 at 19:35