0
$\begingroup$

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

$\endgroup$

closed as off-topic by Namaste, Gabriel Romon, астон вілла олоф мэллбэрг, Daniel W. Farlow, heropup Nov 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.

  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.