How to show that $\lim_{n\to \infty} \frac{a_1 +a_2 + \cdots + a_n}{n} = 0?$ [duplicate]

Solution 1:

For a "better route", use summation by parts with $S_0 = 0$ and $S_k = \sum_{j=1}^k \frac{a_j}{j}$ for $k > 0$ to get

$$\sum_{k=1}^n a_k = \sum_{k=1}^n k\frac{a_k}{k} = \sum_{k=1}^n k(S_k - S_{k-1}) = nS_n - \sum_{k=1}^{n-1}S_k(k+1 - k), $$

and, thus,

$$\frac{1}{n} \sum_{k=1}^n a_k = S_n - \frac{1}{n} \sum_{k=1}^{n-1}S_k$$

Finish by showing that the limit of the RHS is $0$. That $S_n \to S$ implies $n^{-1}\sum_{k=1}^n S_k \to S$ is a well known result.

Solution 2:

I'll prove with your ideas.

Given that $\sum_{k=1}^{\infty}\frac {a_{k}}k$converges, for each $\epsilon$, there exist an $N_{\epsilon}$, such that$$|\sum_{k=N_\epsilon}^{\infty}\frac {a_k}k|<\frac \epsilon2$$for arbitrary $N'>N_\epsilon$, we have $$|\sum_{k=N_\epsilon}^{N'}\frac {a_k}{N'}|<|\sum_{k=N_\epsilon}^{N'}\frac {a_k}k|<\frac \epsilon2$$ while fix this $\epsilon$, $$\forall N_1>\frac {|2{\sum_{k=1}^{N_{}\epsilon}}{a_k}|}{\epsilon},|\frac {\sum_{k=1}^{N_{\epsilon}}{a_k}}{N_1}|<\frac \epsilon2$$ So, set N as max{$N_1,N_{\epsilon,}$}, we can get the conclusion$$\forall n>N,|\frac {\sum_{k=1}^{n}a_n}n|\leq |\frac {\sum_{k=1}^{N_{}\epsilon}{a_k}}n|+|\frac {\sum_{k=N_{}\epsilon}^{n}{a_k}}n|\leq \epsilon$$as the select of $\epsilon$ is arbitrary, the proof is complete.