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$\lim_{n\to \infty}\frac{1+2^2\sqrt{2}+\dots +n^2\sqrt[n]{n}}{n^2(n+1)}$ so i've thought of simplifying into smaller fractions as such $$\lim_{n\to \infty}(\frac{1}{n^2(n+1)}+\frac{2^2\sqrt{2}}{n^2(n+1)}+\dots +\frac{n^2\sqrt[n]{n}}{n^2(n+1)})$$ and as all of the terms we get k/something that approaches infinity, most of the terms approach 0 andon the last one the n^2 cancel out, and we still get all terms to be 0, so the answer is 0? is this logic right? i doubt it's that easy

Xetrez
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    No, your approach is not valid. By that logic, $$ \lim_{n \to \infty} \frac{n}{n} = \lim_{n \to \infty} \overbrace{\frac 1n + \cdots + \frac 1n}^n = \overbrace{0 + \cdots + 0}^n = 0 $$ – Ben Grossmann Dec 07 '20 at 19:09
  • Oh, so how would i approach this? – Xetrez Dec 07 '20 at 19:12

2 Answers2

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@BenGrossmann's comment explains why this doesn't work. Instead use Stolz–Cesàro to write your limit as$$\lim_{n\to\infty}\frac{n^{2+1/n}}{n(3n-1)}=\frac13\color{blue}{\lim_{n\to\infty}n^{1/n}}.$$The blue limit is famously $1$ (see e.g here), making the original limit $\tfrac13$.

J.G.
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An alternative approach: it suffices to show that this limit $L = \lim_{n\to \infty}\frac{1+2^2\sqrt{2}+\dots +n^2\sqrt[n]{n}}{n^2(n+1)}$ is equal to the limit $$ M = \lim_{n \to \infty} \frac{1+2^2+\dots +n^2}{n^2(n+1)}. $$ Note that $M$ is the limit of the left-side Riemann sum $$ \frac{1+2^2+\dots +n^2}{n^2(n+1)} = \sum_{i=0}^{n+1} f(i/n) \cdot \frac 1{n+1}, $$ where $f(x) = x^2$. It follows that $M = \int_0^1 x^2 \,dx = \frac 13$.

Note that $\sqrt[n]{n} \to 1$ as $n \to \infty$ with $\sqrt[n]{n} \geq 1$. It is clear that $L \geq M$, but it remains to be shown that $L \leq M$. To that end, let $\epsilon > 0$. There exists an $N$ such that $n \geq N \implies \sqrt[n] {n}\leq 1 + \epsilon$. For $n > N$, we have $$ \frac{1+2^2\sqrt{2}+\dots +n^2\sqrt[n]{n}}{n^2(n+1)} = \frac 1{n+1}\sum_{i=1}^n \frac{i^2 \sqrt[i]{i}}{n^2}\\ = \frac 1{n+1} \left[ \sum_{i=1}^N \frac{i^2 \sqrt[i]{i}}{n^2} + \sum_{i=N+1}^n \frac{i^2 \sqrt[i]{i}}{n^2}\right] \\ = \frac{C(\epsilon)}{n+1} + \frac 1{n+1} \sum_{i=N+1}^n \frac{i^2 \sqrt[i]{i}}{n^2} \\ \leq \frac{C(\epsilon)}{n+1} + \frac 1{n+1} \sum_{i=N+1}^n \frac{i^2 (1 + \epsilon)}{n^2} \\ = \frac{C(\epsilon)}{n+1} + \frac {1 + \epsilon}{n+1} \sum_{i=N+1}^n \frac{i^2 }{n^2} \\ \leq \frac{C(\epsilon)}{n+1} + \frac {1 + \epsilon}{n+1} \sum_{i=1}^n \frac{i^2 }{n^2}. $$ It follows that $$ L \leq \lim_{n \to \infty} \frac{C(\epsilon)}{n+1} + \frac {1 + \epsilon}{n+1} \sum_{i=1}^n \frac{i^2 }{n^2} \\ = 0 + (1 + \epsilon) M = (1 + \epsilon )M. $$ That is, we have $L \leq (1 + \epsilon)M$ for all $\epsilon > 0$, which means that we must have $L \leq M$, as desired.

Ben Grossmann
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