1. ## Arithmetic Means

For any sequence s_n consider the arithmetic mean

t_n = (s1 + s2 + ... + sn) / n

Prove sn -> s implies tn -> s. Prove there are divergent sequences s_n which in this manner give rise to convergent sequences t_n.

2. ## Re: Arithmetic Means

Originally Posted by veronicak5678
For any sequence s_n consider the arithmetic mean

t_n = (s1 + s2 + ... + sn) / n

Prove sn -> s implies tn -> s. Prove there are divergent sequences s_n which in this manner give rise to convergent sequences t_n.
If $s_{n} \rightarrow s$ we can set each $s_{k}= s + \delta_{k}$ where $\delta_{k} \rightarrow 0$. Now is...

$t_{n} =\sum_{k=1}^{n} \frac{s+\delta_{k}}{n} = s + \sum_{k=1}^{n} \frac{\delta_{k}}{n}$ (1)

... and because each term of the sum at the second term of (1) tends to 0 the sum itself tends to 0...

An example of divergent series that produces a convergent sequence $t_{n}$ is the armonic series. Ibn effect is...

$\sum_{k=1}^{n} \frac{1}{k} \sim \gamma + \ln n$ (2)

... where $\gamma$ is the 'Euler's constant'. From (2) You immediately derive that $t_{n} \rightarrow 0$...

Kind regards

$\chi$ $\sigma$

3. ## Re: Arithmetic Means

Originally Posted by veronicak5678
For any sequence s_n consider the arithmetic mean

t_n = (s1 + s2 + ... + sn) / n

Prove sn -> s implies tn -> s. Prove there are divergent sequences s_n which in this manner give rise to convergent sequences t_n.
Given $\varepsilon>0$, there exists N such that $|s_k-s|<\varepsilon$ whenever k > N. Write $t_n$ as

$t_n-s = \frac{(s_1-s)+(s_2-s)+\ldots+(s_N-s)}n + \frac{(s_{N+1}-s)+(s_{N+2}-s)+\ldots+(s_n-s)}n$,

then show that the first of those fractions tends to 0, and the second fraction stays less than $\varepsilon$, as $n\to\infty.$

For the second part of the question, what about the sequence $s_n=(-1)^n$?