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I have the limit and I tried to find the limit but I am stuck after few steps:

$\lim_{n\to\infty}\dfrac{\log(1^1 +2^2 + \cdots + n^n)}{\sqrt{n^4 + 2n^3\log(n)}-\sqrt{n^4-n^3}} $

$= \lim_{n\to\infty}\dfrac{\log(1^1 +2^2 + \cdots + n^n)(\sqrt{n^4 + 2n^3\log(n)}+\sqrt{n^4-n^3})}{n^4 + 2n^3\log(n)-n^4-n^3} $

$=\lim_{n\to\infty}\dfrac{n^2\log(1^1 +2^2 + \cdots + n^n)\left(\sqrt{1+\frac{ 2\log(n)}{n}}+\sqrt{1-\frac{1}{n}}\right)}{n^3 (2\log(n)-1)}$

$=\lim_{n\to\infty}\dfrac{\log(1^1 +2^2 + \cdots + n^n)\left(\sqrt{1+\frac{ 2\log(n)}{n}}+\sqrt{1-\frac{1}{n}}\right)}{n (2\log(n)-1)} \,\quad? $

What can I do after to find the limit?

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    $\begingroup$ It should be $\sqrt{1+{2\log(n)\over n}}$, not $\sqrt{1+2\log(n)\over n}$. $\endgroup$ – Barry Cipra Feb 8 '19 at 15:29
  • $\begingroup$ "Experimentally" it looks like the answer might be 1, but the convergence is extremely slow. $\endgroup$ – Michael Seifert Feb 8 '19 at 15:30
  • $\begingroup$ You also forgot about an extra minus in front, in the $n^3$ term in the denominator: it should be a +. $\endgroup$ – orion Feb 8 '19 at 23:39
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Try to use the inequality $$ n^{n} < 1^{1} + 2^{2} + \cdots + n^{n} < n^{n} + n^{n} + \cdots + n^{n} = n^{n+1} $$ and apply the squeeze theorem.

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