I want to find the following limit: $$ \lim_{n\rightarrow \infty} \frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2} $$ I guess it converges to 0 and I tried to prove it using the squeeze theorem but it doesn't seem to work. Any ideas?

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    $\begingroup$ What do you mean by $n\rightarrow 0$? $\endgroup$ – Yujie Zha Jan 7 '18 at 23:24
  • $\begingroup$ my mistake, I edited it $\endgroup$ – asdf Jan 7 '18 at 23:29
  • $\begingroup$ I don’t know if it helps, but this can also be written as $$\lim_{n\to\infty}\sum_{x=0}^n\frac{x+1}{(x+n)^2}$$ Sometimes rewriting in sigma notation helps me $\endgroup$ – gen-z ready to perish Jan 8 '18 at 0:16

Your limit can be rewritten as a Riemann sum + $a_n$ where $a_n\rightarrow 0$.

This is a Riemann sum:

\begin{align} \lim_{n\rightarrow \infty} \frac{1}{(n+1)^2}+ \frac{2}{(n+2)^2}+\dots+\frac{n}{(2n)^2}&=\lim_{n\rightarrow \infty} \frac{1}{n}\left ( \frac{1/n}{(1+1/n)^2}+ \frac{2/n}{(1+2/n)^2}+\dots+\frac{n/n}{(1+n/n)^2}\right)\\ &=\int_{0}^1 \frac{x}{(1+x)^2} \text{d}x \\ &=\int_{0}^1 \frac{x+1}{(1+x)^2} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x\\ &=\int_{0}^1 \frac{1}{1+x} \text{d}x -\int_{0}^1 \frac{1}{(1+x)^2} \text{d}x \end{align}

Now, \begin{align}a_n&=\frac{1}{n^2} + \frac{1}{(n+1)^2}+ \frac{1}{(n+2)^2}+\dots+\frac{1}{(2n)^2} \\ &\leq \frac{1}{n^2} + \frac{1}{n^2}+ \frac{1}{n^2}+\dots+\frac{1}{n^2}\\ &=(n+1)\frac{1}{n^2}\rightarrow 0 \end{align}

  • $\begingroup$ Thanks for the answer, although I was looking for something simpler as it adresses to students not familiar with Riemann sums, just basic properties of sequences and limits $\endgroup$ – asdf Jan 8 '18 at 16:05

Note that

$$\frac{1}{n^2} + \frac{2}{(n+1)^2}+ \frac{3}{(n+2)^2}+\dots+\frac{n+1}{(2n)^2}=\sum_{k=n}^{2n}\frac{k-n+1}{k^2} =\sum_{k=n}^{2n}\frac{1}{k}-(n-1)\sum_{k=n}^{2n}\frac{1}{k^2}\sim \log \left(\frac{2n}{n-1}\right)-\frac{n+1}{2n}\to\log 2-\frac12$$

indeed for Harmonic series and Euler–Maclaurin formula

$$\sum_{k=n}^{2n}\frac{1}{k}=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^{n-1}\frac{1}{k}\sim\log2n-\log (2n-1)=\log \left(\frac{2n}{n-1}\right)$$

$$\sum_{k=n}^{2n}\frac{1}{k^2}=\sum_{k=1}^{2n}\frac{1}{k^2}-\sum_{k=1}^{n-1}\frac{1}{k^2}\sim -\frac1{2n}+\frac{1}{n-1}=\frac{n+1}{2n(n-1)}$$


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