A different solution for checking convergence of $a_n=\frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(2n)^2}$

Question

Analyse the convergence of $$a_n=\frac{1}{n^2}+\frac{1}{(n+1)^2}+\dots+\frac{1}{(2n)^2}$$ and find the point of convergence if it does converge.

I proved that it converges to $0$ noting that $$\frac{1}{n^2} \le a_n \le (n+1)\times \frac{1}{n^2}$$ and using Sandwich Theorem.

But I was following a book by Arihant which gives the following solution (and I am quoting it without changing any notations)

Define $a_n=\frac n{(n+n)^2}$ so that $\lim_{n\to \infty} a_n=0$. Now, use Cauchy's Theorem to get $$0=\lim_{n\to \infty} \left(\frac{a_1+\dots a_n}n\right)=\lim_{n\to \infty} \left(\frac{1}{(n+1)^2}+\dots \frac 1{(2n)^2}\right)$$ and hence the result follows.

I believe this solution is absolutely rubbish (although it would be good to have a check). But, I do like the idea and the approach. Can this approach (with necessary modifications) be used to solve these kind of limits?

Answer 1

As it's given that, $a_{n}=\frac{1}{n^2}+\frac{1}{(n+1)^2}+...\frac{1}{(2n)^2}$

To make it more understandable, we can write $a_{n'}$ instead of $a_{n}$ where $n'=0,1,....n$

So, we can write $a_{n'}$ as,

$a_{n'}=\frac{1}{n^2}+\frac{1}{(n+1)^2}+...\frac{1}{(2n)^2}=\sum_{n'=0}^{n}\frac{1}{(n+n')^2}$

$\lim_{n\rightarrow\infty}a_{n'}$=$\lim_{n\rightarrow\infty}(\sum_{n'=0}^{n}\frac{1}{(n+n')^2}$)

From here, we can easily interpret that,

$\lim_{n\rightarrow\infty}(\sum_{n'=0}^{n}\frac{1}{(n+n')^2}$)=0

So, $\lim_{n\rightarrow\infty}a_{n'}$=0.

Answer 2

just for the fun of it :-) \begin{align} 0\leq a_n &= \sum_{j=0}^{n}\frac{1}{(n+j)^2} \\ &\leq \sum_{j=0}^{n}\frac{1}{(n+j)n}\\ & = \frac{1}{n^2}\sum_{j=0}^{n}\frac{1}{1+\frac jn}\\ & = \lim_{n\to \infty} \frac1n\times\int_0^1\frac{dx}{1+x}\\ & = \lim_{n\to \infty} \frac1n \times \ln2\\ & = 0 \end{align}

Answer 3

Compute an upper bound $$ \begin{align} \sum_{k=n}^{2n}\frac1{k^2} &\le\sum_{k=n}^{2n}\frac1{(k-1)k}\tag{1a}\\ &=\sum_{k=n}^{2n}\left(\frac1{k-1}-\frac1k\right)\tag{1b}\\ &=\frac1{n-1}-\frac1{2n}\tag{1c}\\ &=\frac{n+1}{2n(n-1)}\tag{1d}\\ &\le\frac1{2n-4}\tag{1e} \end{align} $$ Explanation:
$\text{(1a):}$ $\frac1{k^2}\le\frac1{(k-1)k}$
$\text{(1b):}$ partial fractions
$\text{(1c):}$ telescoping series
$\text{(1d):}$ combine fractions
$\text{(1e):}$ cross multiply: $2n^2-2n-4\le2n^2-2n$

Although $0$ is a sufficient lower bound, we can compute a better lower bound $$ \begin{align} \sum_{k=n}^{2n}\frac1{k^2} &\ge\sum_{k=n}^{2n}\frac1{k(k+1)}\tag{2a}\\ &=\sum_{k=n}^{2n}\left(\frac1k-\frac1{k+1}\right)\tag{2b}\\ &=\frac1n-\frac1{2n+1}\tag{2c}\\ &=\frac{n+1}{n(2n+1)}\tag{2d}\\ &\ge\frac1{2n}\tag{2e} \end{align} $$ Explanation:
$\text{(2a):}$ $\frac1{k^2}\ge\frac1{k(k+1)}$
$\text{(2b):}$ partial fractions
$\text{(2c):}$ telescoping series
$\text{(2d):}$ combine fractions
$\text{(2e):}$ cross multiply: $2n^2+2n\ge2n^2+n$

Therefore, $$ \frac1{2n}\le\sum_{k=n}^{2n}\frac1{k^2}\le\frac1{2n-4}\tag3 $$ and we can apply the Squeeze Theorem.

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Of course, if we know that the series $\sum\limits_{k=1}^\infty\frac1{k^2}$ converges, then the desired result follows from the fact that the partial sums of a convergent series are a Cauchy sequence.