Compute $\lim_{n \rightarrow \infty} (\frac{2^{1/n}}{n+1} + \frac{2^{2/n}}{n+1/2} +…+ \frac{2^{n/n}}{n+1/n})$

Question

Its weird as I have a solution but neither I'm satisfied by my method nor I can find why is this wrong.
We write the sum after some simple steps as $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{r2^{r/n}}{r+1/n}$$ Now we can say that $\frac{r}{r+1/n} \rightarrow 1$ as $n \rightarrow \infty$
Thus we have the riemann sum $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n} 2^{r/n}$$ which when solved gives the value $$\int_0^1 2^x dx = \frac{1}{\ln 2}$$

Answer 1

Noting $$ n<n+\frac 1k <n+1$$ one has $$ \sum_{k=1}^n\frac{2^{k/n}}{n+1}<\sum_{k=1}^n\frac{2^{k/n}}{n+\frac1k}<\sum_{k=1}^n\frac{2^{k/n}}{n}. $$ Since $$ \lim_{n\to\infty}\sum_{k=1}^n\frac{2^{k/n}}{n+1}=\lim_{n\to\infty}\frac n{n+1}\frac1n\sum_{k=1}^n2^{k/n}=\int_0^12^xdx=\frac1{\ln2}$$ and $$ \lim_{n\to\infty}\sum_{k=1}^n\frac{2^{k/n}}{n}=\lim_{n\to\infty}\frac1n\sum_{k=1}^n2^{k/n}=\int_0^12^xdx=\frac1{\ln2}$$ so $$ \lim_{n\to\infty}\sum_{k=1}^n\frac{2^{k/n}}{n+\frac1k}=\frac1{\ln2}. $$