I am working on the following problem.
Let $a_n = \frac{1+2^{-2}+…+n^{-2}}{n}$ for $n\in \mathbb{N}$.Then I have to check the convergence of both the sequence $(a_n)$ and the series $\sum a_n$.
My attempt: Let $\alpha_n = \frac{1}{n^2}$ then using this result if $\lim_{n\rightarrow\infty}\alpha_n=L$ then $\lim_{n\rightarrow\infty} \frac{\alpha_1+\alpha_2+\cdots+\alpha_n}n=L$, I observe that the sequence $(a_n)$ convergese as the limit of sequence $a_n = \frac{1+2^{-2}+…+n^{-2}}{n}$ is zero as $n\rightarrow \infty$. Am I on the right path? How to check the convergence of the series $\sum a_n$. I am stucked here.
Thanks for your assistance
Best Answer
Notice that $$\sum_{k=1}^nk^{-2}=1+\sum_{k=2}^nk^{-2}=1+\sum_{k=2}^n\int\unicode{120793}\{k-1<x<k\}k^{-2}\mathrm dx\le1+\int_1^nx^{-2}\mathrm dx=1+1-n^{-1},$$ so we have $a_n\le 2/n$ and hence $\lim_{n\rightarrow\infty}a_n=0$. On the other hand, we have $a_n\ge 1/n$ and hence $$\sum_{n=1}^\infty a_n\ge\sum_{n=1}^\infty\frac{1}{n}=\infty,$$ since the latter sum is the harmonic series. To be more precise, using another integral bound, we have $$\sum_{k=1}^n\frac{1}{k}=\sum_{k=1}^n\int\unicode{120793}\{k<x<k+1\}\frac1k\mathrm dx\ge\int_1^{n+1}\frac1x\mathrm dx=\ln(n+1).$$ In particular, this shows that $\sum_{n=1}^\infty\frac{1}{n}=\infty$.