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, heropup Nov 24 '16 at 5:04

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  • $\begingroup$ Write it as a Riemann sum? $\endgroup$ – Jack Nov 21 '16 at 16:14
  • 1
    $\begingroup$ @Jack No thanks. $\endgroup$ – 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.

  • $\begingroup$ if k=2 Starts How can? $\endgroup$ – Expal1975 Nov 21 '16 at 17:28
  • $\begingroup$ I don't understand. What do you mean? $\endgroup$ – Clement C. Nov 21 '16 at 17:28
  • $\begingroup$ I edit . look . k=2 $\endgroup$ – Expal1975 Nov 21 '16 at 17:31
  • $\begingroup$ The same argument will apply, try to do so and you'll see. (That's a good way to check you understand the answer.) Also, in the future try to refrain from changing your question after it has been answered. $\endgroup$ – Clement C. Nov 21 '16 at 17:33
  • $\begingroup$ $$n \leq \log\sum_{k=1}^{2^n} k^{\frac{1}{n^2}} \leq n+\frac{1}{n}.$$ 1 $$1\leq\frac{1}{n} \log\sum_{k=1}^{2^n} k^{\frac{1}{n^2}} \leq n+\frac{2}{n}$$ $$ \lim_{ x \to \infty }\frac{1}{n} \log\sum_{k=1}^{2^n} k^{\frac{1}{n^2}}=?$$ $\endgroup$ – Expal1975 Nov 21 '16 at 17:39

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