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I am doing a guided problem to find the relationship between the convergence of a sequence $\{s_n\}$ of real numbers, and the convergence of its arithmetic mean $\sigma_n := (s_1 + \cdots + s_n) /n$.

It is not hard to see that $s_n \longrightarrow s$ then $\sigma_n \longrightarrow s$, but the converse is false.

However, if we introduce $a_1 = s_1$ and $a_n = s_n - s_{n-1}$ for $n > 1$, the exercise is to prove that if $n a_n \longrightarrow 0$ and $\sigma_n \longrightarrow s$ then also $s_n \longrightarrow s$, which is some sort of converse, and which I cannot prove.

There is a hint, to prove that $s_n - \sigma_n = \frac{1}{n} \sum_{j=1}^{n}{(j-1)a_j}$, which is clear. I am trying to prove that this goes $0$, in which case we would be done by $s_n = (s_n - \sigma_n) + \sigma_n \longrightarrow 0 + s = s$.

Is this the right way to do it ? How to prove that the sum $\frac{1}{n} \sum_{j=1}^{n}{(j-1)a_j} \rightarrow 0$ ?

Thanks !

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up vote 1 down vote accepted

You have to use the extra condition $n\,a_n\to0$.

Deduce first that $(n-1)a_n\to0$. Then applying to the sequence $(n-1)a_n$ the result about the convergence of the arithmetic mean, you get the desired result.

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OMG that's beautiful. Thanks. – Bogdan Oct 16 '12 at 20:25

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