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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

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  • $\begingroup$ 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
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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.

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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.

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