Finding $\lim_{n \to \infty}\left(\frac{2^{\frac{1}{n}}}{n+1}+\frac{2^{\frac{2}{n}}}{n+\frac{1}{2}}+\cdots+\frac{2^{\frac{n}{n}}}{n+\frac1n}\right)$

Question

I am trying to determine

$$\lim_{n \to \infty}\left(\frac{2^{\frac{1}{n}}}{n+1}+\frac{2^{\frac{2}{n}}}{n+\frac{1}{2}}+\cdots+\frac{2^{\frac{n}{n}}}{n+\frac1n}\right)$$

Heres is my attempt: I know that if $\lim a_n=:a$, and $\lim b_n = :b$ we get that $\lim (a_n+b_n)=a+b$

Obviously

$$\lim_{n \rightarrow \infty} \frac{2^{\frac{1}{2}}}{n+1}=0$$

$$\lim_{n \rightarrow \infty} \frac{2^{\frac{2}{n}}}{n+\frac{1}{2}}=0$$

The only thing left is to determine

$$\lim_{n \rightarrow \infty} \frac{2^{\frac{n}{n}}}{n+\frac{1}{n}}=0$$

Thus we get that $$\lim_{n \to \infty}\left(\frac{2^{\frac{1}{n}}}{n+1}+\frac{2^{\frac{2}{n}}}{n+\frac{1}{2}}+\cdots+\frac{2^{\frac{n}{n}}}{n+\frac1n}\right) = 0$$

My question is if my approach is correct. Especially because this exercise was labled "very difficult", which it really does not seem like.

Answer 1

\begin{align} \lim_{n\to\infty}\sum_{j=1}^n\frac{2^{\frac jn}}{n+\frac 1j} &=\lim_{n\to \infty}\frac1n \sum_{j=1}^n\frac{2^{\frac jn}}{1+\frac 1{jn}}\\ &=\int_0^12^xdx \end{align}

Answer 2

We have that

$$\sum_{j=1}^n\frac{2^{\frac jn}}{n+\frac 1j}=\frac1n\sum_{j=1}^n\frac{2^{\frac jn}}{1+\frac 1{jn}}\to \frac1{\log 2}$$

indeed

$$\frac{2^{\frac jn}}{1+\frac 1{n}}\le \frac{2^{\frac jn}}{1+\frac 1{jn}}\le \frac{2^{\frac jn}}{1+\frac 1{n^2}}$$

$$\frac1{n+1}\sum_{j=1}^n2^{\frac jn}\le \frac1n\sum_{j=1}^n\frac{2^{\frac jn}}{1+\frac 1{jn}}\le \frac 1{n+\frac1n}\sum_{j=1}^n2^{\frac jn}$$

and by geometric series $\sum_{j=1}^n2^{\frac jn}=\frac{2^\frac1n}{2^\frac1n-1}$, thus

$$\frac {n2^\frac1n}{n+1}\frac{\frac1n}{2^\frac1n-1}\le \frac1n\sum_{j=1}^n\frac{2^{\frac jn}}{1+\frac 1{jn}}\le \frac {n2^\frac1n}{n+\frac1n}\frac{\frac1n}{2^\frac1n-1}$$

with $$\frac{\frac 1n}{2^\frac1n-1} \to \frac1{\log 2}$$