Given any convergent series $(a_n)_{n \in \mathbb{N}}$,consider a new sequence $(b_n)_{n \in \mathbb{N}}$, defined as $$b_n = \frac{1}{n}(a_1 + a_2 + \dots+a_n)=\frac{1}{n}\sum_{k=1}^na_k $$ for every $n \in \mathbb{N}$ ($b_n$ is the arithmetic mean of the first $n$ elements in $a_n$).

How do I proof that $(b_n)_{n \in \mathbb{N}}$ is convergent with the same limit?

I have tried using the fact that $b_n \leqslant \text{sup}\{a_m\vert m \geqslant n\}$ and $b_n \geqslant \text{inf}\{a_m\vert m \geqslant n\}$, and subsequently taking limits and using that $L= \text{limsup}_{n \to \infty}a_n =\text{liminf}_{n \to \infty}a_n$. Is this sound?


Since $\{a_n\}$ is convergent, it is bounded; let $A=\sup_n |a_n|$. Choose $N_1$ so that $|a_n-a|<\frac\varepsilon2$ for $n\geqslant N_1$, and $N\geqslant\frac{4N_1A}\varepsilon$. Then for $n\geqslant N_1$ we have \begin{align} |b_n - a| &= \left|\sum_{k=1}^n \frac{a_k-a}n\right|\\ &\leqslant \frac1n\sum_{k=1}^n|a_k-a|\\ &= \frac 1n\sum_{k=1}^{N_1}|a_k-a| + \frac1n\sum_{k=N_1+1}^n|a_k-a|\\ &\leqslant \frac{2N_1A}n + \frac{\varepsilon(n-N_1)}{2n}\\ &\leqslant \frac\varepsilon2 + \frac\varepsilon2\\ &=\varepsilon. \end{align}



Suppose $(a_n)$ converge to $\ell$.

$$\left|b_n-\ell\right|=\left|\frac{1}{n}\sum_{k=1}^n a_k-\ell\right|=\frac{1}{n}\left|\sum_{k=1}^n(a_k-\ell)\right|\leq \frac{1}{n}\sum_{k=1}^n|a_k-\ell|.$$

Using the the fact that $a_n\longrightarrow \ell$, you can easily conclude.


This is a consequence of the theorem of Cesaro-Stolz, the discrete version of the l'Hopital rule. Set $c_n=1$, then since $a_n/c_n$ converges,
$$ \frac{a_1+a_2+…+a_n}{c_1+c_2+…+c_n} $$ also converges, and has the same limit. The answer of Math1000 contains the standard proof (strategy) of that statement, which can also be found in wikipedia or linked there.

See also Stolz-Cesàro Theorem for some in-depth discussion of the theorem and its geoemtry.

  • $\begingroup$ Can it also be proven using the limsup/liminf argument I made? $\endgroup$ – Michael Angelo Jan 28 '16 at 13:32
  • 2
    $\begingroup$ No, I do not see why those statements should be true. The start of a converging sequence can be quite arbitrary. You can have, as example, 10000 terms of $a_n=2$ and the remaining infinity of terms equal 1, so that $b_{10001}>1.5$ and $\sup{n\ge 10001}a_n=1$. $\endgroup$ – LutzL Jan 28 '16 at 13:41

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