Find limit of $$a_n = \sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n + \frac{1}{k}}$$
My attempt is:
$$\sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n + \frac{1}{n}} < \sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n + \frac{1}{k}} < \sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n}$$
Note as $n \to \infty \Rightarrow n + \frac{1}{n} \to n \Rightarrow \sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n + \frac{1}{n}} \to \sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n}$.
Therefore, by squeeze theorem the limit of central sum is equal to the limit of lateral sums.
Then let's find limit when $n \to \infty$ of the following $\sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n} = 2^{\frac{1}{n}} \sum \limits_{k = 1}^n \frac{1}{n} \cdot2^k$.
Let's take $f(x) = 2^x$, the norm of partition is equal to $\frac{1}{n} \to 0$, $\xi_k = \frac{k}{n}$, then $\sum \limits_{k = 1}^n \frac{1}{n} \cdot2^{\frac{k}{n}} \to \int \limits_0^1 2^x dx = \frac{2^x}{\ln 2}\Big|_0^{1} = \frac{1}{\ln2}$.
So, our initial sum has the following limit.
I have a few questions about my solution:
I'm pretty sure that my explanation of why $\sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n + \frac{1}{n}} \to \sum \limits_{k = 1}^n \frac{2^{\frac{k}{n}}}{n}$ is not strict enough. Can someone please explain this moment more carefully?
Is it true that $n + \frac{1}{n} \to n$?
In general, is the following is true: if $ f(x) \to g(x) \Rightarrow g(x) \to f(x)$.
Are there any other options to solve this particular problem using Riemann sum?
It is a good idea of user23316192 to use the squeeze theorem.
We can improve his reasonings a little by simplifying the estimation of the lateral sums:
$$\sum _{k=1}^n \frac{2^{k/n}}{n+1} < a_n < \sum _{k=1}^n \frac{2^{k/n}}{n} $$
The main part of these lateral sums is the geometric sum
$$\sum _{k=1}^n 2^{k/n}=\frac{2^{1/n}}{2^{1/n}-1}$$
The limit to be considered is (for $k = 0$ or $k = 1$)
$$\lim_{n\to \infty } \, \frac{2^{1/n}}{\left(2^{1/n}-1\right) (k+n)}=\frac{1}{\log (2)}$$
Main details are here the definition $2^x = e^{x \log 2}$ and the power series development $e^x \simeq 1 + x$ from which we find
$$\frac{2^{1/n}}{2^{1/n}-1}=\frac{1}{1-2^{-\frac{1}{n}}}=\frac{1}{1-e^{-\frac{\log (2)}{n}}}\simeq \frac{1}{1-\left(1-\frac{\log (2)}{n}\right)}=\frac{1}{\frac{\log (2)}{n}}$$
hence finally
$$\lim_{n\to \infty } \, a_n =\frac{1}{\log (2)}$$