I'm working on a problem set which was given by our analysis lecturer

(a) Let $(a_n)$ be a convergent sequence with limit $a$. show that the arithmetic mean $$s_n := \frac{1}{n} \sum_{k=1}^n a_k$$ of the sequence $(a_n)$ converges also to $a$

(b) Give a divergent sequence which has a converging arithmetic mean

My Attempt:

(a) As we know the definition for converging sequences: $$\forall \epsilon > 0 \ \exists N(\epsilon) \in \mathbb{N} \ \text{such that} \ \forall n > N \vert a_n - a \vert < \epsilon$$ now we want to show that the arithmetic mean is also converging towards $a$ therefore we can take the limit of $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n a_k$$ further we say that $\forall a_n$ where $n > N$ we know $ a_n \rightarrow a$ this means: $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=N}^n a_k \Leftrightarrow \frac{\overbrace{a_N + \ldots + a_n}^{= na}}{n \rightarrow \infty} \rightarrow a$$ the part where we sum from $1$ to $N$ we can neglect because we only observe for the converging range (correct?)

(b) as we know $ a_n = (-1)^n$ is divergent ($\rightarrow (-1) \ \text{and} \ 1$). Hence arithmetic mean is convergent: $$ \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=N}^n a_k \Leftrightarrow \frac{ \overbrace{-1 +1 -1 +1 \ldots}^{= 0}}{n} = 0$$

I am thankful for any hint and advice

  • $\begingroup$ What is a "converging range"? - Also, your counterexample is fine, but your analysis of it is not $\endgroup$ – Hagen von Eitzen Oct 30 '14 at 14:58
  • $\begingroup$ @HagenvonEitzen what do you mean by that? $\endgroup$ – Mainviel Oct 30 '14 at 15:02

We aim to show $s_n= \frac{1}{n} \sum_{k=1}^n a_k-a\to 0 \iff \frac{1}{n} \sum_{k=1}^n (a_k-a)\to 0$

Hence it suffices to show the case when $s=0$, that is $a_n\to 0 \implies s_n\to 0$

That is $\forall \epsilon > 0 \ \exists N_1 \in \mathbb{N} \ \text{such that} \ \forall n > N_1 \vert a_n\vert < \epsilon$

Let $M=\sum_{i=0}^N |a_i|, \exists N_2 \in \mathbb{N} \ \frac{M}{N_2} < \epsilon$

Then $\forall n>N=\max(N_1,N_2), |s_N|\leq\frac{\sum_{i=0}^{n} |a_i|}{n}=\frac{M+\sum_{i=N+1}^{n} |a_i|}{n}<\frac{M}{n}+\frac{(n-N)\epsilon}{n}< 2\epsilon$

Hence we have done.

For the second question, just note $s_n=-1$ for odd $n$, $s_n=0$ for even $n$. Other things you are correct.


a) We can't entirely neglect the part from 1 to $N$ because it's part of the mean. But for each $\varepsilon$ the part from 1 to $N$ contributes a constant to the numerator and a constant to the denominator. You can show, then, that after sufficiently many terms (how many?), the mean is within $2\varepsilon$ of the limit. Then you are done because $2\varepsilon \rightarrow 0$ as $\varepsilon \rightarrow 0$.

b) You chose a valid example but your proof isn't completely valid. Specifically, the numerator alternates between 0 and -1 so it isn't always 0. Thus the mean alternates between 0 and $-\frac{1}{n}$, which clearly is a convergent sequence.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.