# Simple series proof

1. Nov 6, 2013

### synkk

show that if $a_n$ is any sequence and m is any integer such that the series $\displaystyle \sum_{n=m}^\infty a_n$ converges, then $\displaystyle\lim_{k \to \infty} \sum_{n=k}^{\infty } a_n = 0$

let $S_N = \displaystyle \sum_{n=m}^N a_n$

$\lim_{N \to \infty} S_N = l$ from definition
also,
$\lim_{N \to \infty} S_{N-1} = l$

$S_N - S_{N-1} = a_N$

$\lim_{N \to \infty} (S_N - S_{N-1}) = \lim_{N \to \infty}(S_N) - \lim_{N \to \infty} S_{N-1} = 0$ then $\lim_{N \to \infty} a_N = 0$ i.e. $\lim_{N \to \infty} \sum_{n=m}^N a_n = 0 \Rightarrow \lim_{k \to \infty} \sum_{n=k}^{\infty} a_n = 0$

as you can see I don't really know how to prove this statement, but I have attempted it.

Could someone show me how I can proceed to do the proof properly?

2. Nov 6, 2013

### pasmith

You are given that
$$\sum_{n=m}^\infty a_n$$
converges. Thus you have
$$L = \sum_{n=m}^\infty a_n = S_{k-1} + \sum_{n = k}^{\infty} a_n$$
so that
$$|L - S_{k-1}| = \left| \sum_{n = k}^{\infty} a_n \right|$$

3. Nov 7, 2013

### synkk

how did you get $L = \sum_{n=m}^\infty a_n = S_{k-1} + \sum_{n=k}^{\infty} a_n$?

4. Nov 7, 2013

### FeDeX_LaTeX

They've just written it as a sum of two series.

5. Nov 7, 2013

### JorisL

I do not believe it is correct though.
It should read $L=\sum_{n=0}^\infty a_n = S_{k-1}+\sum_{n=k}^\infty a_n$.
This is to be seen from the definition of $S_k = \sum_{n=0}^k a_n$.

The final step in post #2 is correct. (Although I'm not entirely sure why the absolute value shows up, but it doesn't make any difference here)

6. Nov 7, 2013

### pasmith

Read the OP. We are told that $\sum_{n=m}^\infty a_n$ converges for some integer $m$. In particular, we are not told that $m$ is positive. Therefore the OP's definition of $S_k = \sum_{n=m}^{k} a_n$ is correct.

However, in any event the behaviour of a finite number of terms at the beginning of a sequence does not affect convergence of the series.

7. Nov 7, 2013

### JorisL

My bad, sorry for the confusion.
But this doesn't change anything indeed.