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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Gabriel Romon, астон вілла олоф мэллбэрг, Daniel W. Farlow, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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