# Mean of the terms of convergent sequence [duplicate]

Consider a convergent sequence $a_1,a_2,a_3\cdots a_n$ tending to a limit A. Now consider the sequence $K_1,K_2,K_3 \cdots K_n$ such that $K_n =\cfrac {a_1+a_2+...a_n}n$. Now what I guess is that as an tends to infinity $K_n$ tends to $a_n$. What I can say is that the sequence $K_n$ is a bounded one. As such it must have a convergent sub sequence. Is this sequence also convergent.

I tried to resolve this by first trying to apply squeeze theorem as $K_n$ always lies between the minimum and maximum of that sequence. But I could not meet up rigor or any clear result.

Next I tried to apply Cauchy's principle, but here I don't think that one can reach to a strong consequence except few suggestive apprehensions.

I hope someone can guide me on this problem with a proof.

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Moyenne de Cesàro. –  user31280 Oct 27 '12 at 5:21
See also: Prove convergence of a sequence. There and in linked questions several proofs (or several versions of the same proof) can be found. AFAIK the proofs I've seen are not similar to the proof of Stolz-Cesaro Theorem. The question limit of quotient of two series has relatively detailed proof of Stolz-Cesaro, also the question Stolz-Cesaro Theorem has several links. –  Martin Sleziak Oct 27 '12 at 6:20

## marked as duplicate by Did, T. Bongers, mrf, TZakrevskiy, azimutAug 26 '13 at 7:46

They only coincide when $a_n$ is a convergent sequence. But $K_n$ can converge even though $a_n$ doesn't converge. Take for example $a_n=(-1)^n$, then $|K_n| \leq \frac{1}{n}$ and hence $K_n$ converges $0$. –  Dominic Michaelis Aug 26 '13 at 5:35