# how to solve $\lim_{n \rightarrow \infty } \frac{1}{n} \log \left(\sum_{k=2}^{2^n} k^{\frac{1}{n^{2}}}\right)$ [closed]

How to evaluate the following limit?

$$\lim_{n \rightarrow \infty } \frac{1}{n} \log \left(\sum_{k=2}^{2^n} k^{\frac{1}{n^2}}\right)$$ enter image description here

## closed as off-topic by Namaste, Gabriel Romon, астон вілла олоф мэллбэрг, Daniel W. Farlow, heropupNov 24 '16 at 5:04

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• Write it as a Riemann sum? – Jack Nov 21 '16 at 16:14
• @Jack No thanks. – Did Nov 21 '16 at 17:21

For any $1\leq k\leq 2^n$, we have $$1\leq k^{\frac{1}{n^2}} \leq 2^{\frac{1}{n}}$$ so that $$2^n \leq \sum_{k=1}^{2^n} k^{\frac{1}{n^2}} \leq 2^n \cdot 2^{\frac{1}{n}} = 2^{n+\frac{1}{n}}.$$ Taking the logarithm (I assume in base 2 (?)), $$n \leq \log\sum_{k=1}^{2^n} k^{\frac{1}{n^2}} \leq n+\frac{1}{n}.$$ and you can conclude by the squeeze theorem.
• $$n \leq \log\sum_{k=1}^{2^n} k^{\frac{1}{n^2}} \leq n+\frac{1}{n}.$$ 1 $$1\leq\frac{1}{n} \log\sum_{k=1}^{2^n} k^{\frac{1}{n^2}} \leq n+\frac{2}{n}$$ $$\lim_{ x \to \infty }\frac{1}{n} \log\sum_{k=1}^{2^n} k^{\frac{1}{n^2}}=?$$ – Expal1975 Nov 21 '16 at 17:39