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?
 A: 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
$$
\frac{n}{n+1} \leq \frac{\sum\limits_{k=1}^n \frac{2^{k/n}}{n+\frac{1}{k}}}{\sum\limits_{k=1}^n \frac{2^{k/n}}{n}} < 1
$$
A: Let $f(x)$
be a function such that,
for $x \ge 0$,
$f(x) \ge 0$
and
$f(x)$ is bounded for
$0 < x \le 1$.
Let
$M = \max_{0 < x \le 1} f(x)
$.
This problem is the case
$f(x) = 2^x$,
so $M = 2$.
Let
$g(n)
=\sum\limits_{k=1}^n \frac{f(k/n)}{n+\frac{1}{k}} 
= \frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{1 + \frac{1}{kn}}
$.
Since
$1 \ge \frac1{1+x}
\ge 1-x
$
for
$0 \le x \le 1$,
$g(n)
\le \frac{1}{n} \sum\limits_{k=1}^n f(k/n)
$
and
$\begin{array}\\
g(n)
&\ge \frac{1}{n} \sum\limits_{k=1}^n f(k/n)(1 - \frac{1}{kn})\\
&= \frac{1}{n} \sum\limits_{k=1}^n f(k/n) - \frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{kn}\\
&= \frac{1}{n} \sum\limits_{k=1}^n f(k/n) - \frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{k^2}\\
&\ge \frac{1}{n} \sum\limits_{k=1}^n f(k/n) - \frac{f(1)}{n} \sum\limits_{k=1}^n \frac{1}{k^2}\\
&\ge \frac{1}{n} \sum\limits_{k=1}^n f(k/n) - \frac{2M}{n}
\qquad\text{since }  \sum\limits_{k=1}^n \frac{1}{k^2}
\lt 1+\sum\limits_{k=2}^n \frac{1}{k(k-1)} < 2\\
\end{array}
$
Therefore
$0
\le \frac{1}{n} \sum\limits_{k=1}^n f(k/n)
-\frac{1}{n} \sum\limits_{k=1}^n \frac{f(k/n)}{1 - \frac{1}{kn}}
\le \frac{2M}{n}
$.
Letting $n \to \infty$,
we see that the limits
of the two sums
are the same.
Note that the function
inside being $2^{k/n}$
does not matter,
only that it is bounded.
