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Suppose $\lim_{j\to\infty} a_j=L$

then show that

$$\lim_{j \rightarrow \infty}\frac{a_1+a_2+a_3+\cdots+a_j}{j}=L$$

How would you show this I am entirely certain.

Is this because when you add up all your terms in the sequence and divide by how many you have you have an average, and the limit goes to this "average"

Michael Hardy
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Fernando Martinez
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  • http://www.sosmath.com/calculus/sequence/hardlim/hardlim.html – R.N Sep 20 '15 at 19:08
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    http://math.stackexchange.com/questions/207910/prove-convergence-of-the-sequence-z-1z-2-cdots-z-n-n-of-cesaro-means – R.N Sep 20 '15 at 19:09

1 Answers1

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Given $\varepsilon>0$, one can find $N$ so large that whenever $n > N$ then $|a_n -L| < \dfrac \varepsilon 2$.

Then \begin{align} & \frac{a_1 + \cdots + a_N + a_{N+1} + \cdots + a_M} M \\[10pt] = {} & \frac N M \cdot \underbrace{\frac{a_1+\cdots+a_N} N}_\text{first average} {}+{} \frac{M-N} M \cdot \underbrace{\frac{a_{N+1} + \cdots + a_M}{M-N}}_\text{second average}. \end{align} The second average above is the average of numbers between $L\pm\dfrac\varepsilon2$, and is thus itself between those bounds.

As $M$ grows while $N$ stays fixed, we have $\dfrac N M\to 0$ so the whole term involving the first average approaches $0$, and $\dfrac{M-N} M \to 1$.

For large enough $M$, see if you can show the whole thing must therefore be between $L\pm\varepsilon$.

Michael Hardy
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