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

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?

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

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.