<s>[INST] How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$? How to show that $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}}$ is like $\sum_{k=1}^n\frac{2^{\frac{k}{n}}}{n}$ if $n\rightarrow\infty$?
I tried show that difference between terms of sums $\rightarrow 0$
$|\frac{2^{\frac{k}{n}}}{n+\frac{1}{k}} - \frac{2^{\frac{k}{n}}}{n}| = 2^\frac{k}{n}(\frac{1}{n+\frac{1}{k}} - \frac{1}{n})= 2^\frac{k}{n}(\frac{n-n-\frac{1}{k}}{n^2+\frac{n}{k}}) = -2^\frac{k}{n} \frac{1}{n^2k+n}\rightarrow 0$ if $n\rightarrow \infty$
But i'm not sure that is right method, because difference of terms can accumulete and it would be $\neq 0$
Which correct method i must use to prove it?
 [/INST] If you're trying to show the difference goes to $0$, here's a big hint:
$$
\sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}} = \frac{1}{n} \sum\limits_{k=1}^n \frac{2^{k/n}}{1 + \frac{1}{kn}}
$$
and hence
$$ 0 \leq 
 \sum\limits_{k=1}^n \frac{2^{k/n}}{n} - \sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}}= \frac{1}{n}\sum\limits_{k=1}^n\frac{(1+\frac{1}{kn})2^{k/n} - 2^{k/n}}{1 + \frac{1}{kn}} = \frac{1}{n}\sum\limits_{k=1}^n\frac{2^{k/n}}{kn\left(1 + \frac{1}{kn}\right)} < \frac{1}{n} \sum\limits_{k=1}^n \frac{2}{k^2}
$$
Or, perhaps the following will serve you better:
$$
\sum_{k=1}^n \frac{2^{k/n}}{n + 1} \leq \sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}} < \sum_{k=1}^n \frac{2^{k/n}}{n}
$$
and hence
$$
\frac{n}{n+1} \leq \frac{\sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}}}{\sum\limits_{k=1}^n \frac{2^{k/n}}{n}} < 1
$$
</s>