# Divergent succession, but with convergent sum average.

An example of a sequence $a_n$ such that: $$a_n\rightarrow\pm\infty$$ but $$b_n=\frac{\sum_{k=1}^{n}a_k}{n}$$ converge.

Assume without loss of generality that $a_n > 0$ always. Suppose that $b_n \to L$ a finite limit. Since $a_n \to \infty$, there exists some $N$ such that for all $n \geq N$, we have that $a_n > 2L$. But then for $n > 2N$, $$\sum_{k = 1}^{n} a_n > 0 + \sum_{k = N}^{n} 2L > Ln.$$ So in particular, the average of these terms will always be more than $L$, contrary to the claim. This is a contradiction. $\diamondsuit$
More generally, it's true that if $a_n \to L$, then the average value of $a_n$ also converges to $L$.
To match the edit, consider $$a_n = (-1)^n \log n.$$
• If $n > 2N$, then we have at least $N$ terms at least $2L$ in size. – davidlowryduda Apr 20 '15 at 7:12
• that is true if $a_n>0$, but if $a_n=(-1)^nn^\alpha$ with $\alpha$ suitable ? – Alex Med Apr 20 '15 at 7:12
• @ShoumanDas I don't know what you mean. The sum of the first $N$ terms of the sequence $\sqrt n$ is about $N^{3/2}$. So the average of the first $N$ terms is about $\sqrt N$, which clearly goes to $\infty$. – davidlowryduda Apr 20 '15 at 7:20
• Sure. Take something like $(-1)^n\log n$. Roughly speaking, since $\log (n+1) - \log n \approx 0$ for large $n$, you'll get $b_n \to 0$. – davidlowryduda Apr 20 '15 at 7:30