I'm asked to prove
$$\displaystyle\lim_{n\to \infty} \sum_{k=0}^{n} 2^{\frac{-kn}{k+n}}=2$$
Define $$b_{k,n} = \begin{cases}
0,& n \le k\\
2^{\frac{-kn}{k+n}},& n \ge k\\
\end{cases}$$
Our limit is equivalent to finding $\displaystyle\lim_{n\to \infty} \sum_{k=0}^{\infty} b_{k,n}$. It's appealing to write
$$\displaystyle\lim_{n\to \infty} \sum_{k=0}^{\infty} b_{k,n}=^?\sum_{k=0}^{\infty}\lim_{n\to \infty} b_{k,n}=\sum_{k=0}^{\infty} 2^{-k}=2$$
Question 1:
Can we justify the swapping of sum and limit here? (solved)
Question 2:
Is there any alternative approach?
Best Answer
The series can be dominated by $$ b_{k,n} \le 2^{-\frac{k}{2}}$$ For $n < k$ this is obivously true, as we have $b_{k,n} = 0$. For $n\ge k$ we have $$ n \ge k$$ $$ \frac1n \le \frac 1k$$ $$ \frac1k + \frac1n \le \frac2k$$ $$ \frac{kn}{k+n} = \frac{1}{\frac1k + \frac1n} \ge \frac{k}{2}$$ $$ b_{k,n} = 2^{-\frac{kn}{k+n}} \le 2^{-\frac{k}{2}}$$ Series $ \sum_{k=0}^\infty 2^{-\frac{k}{2}} = \sum_{k=0}^\infty \frac{1}{(\sqrt{2})^k}$ is convergent, so from the dominated convergence theorem, you can justify the swapping of the sum and the limit.
More fundamental proof: Let us decompose $$ \sum_{k=0}^n 2^{-\frac{kn}{k+n}} = \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} + \sum_{k={\lfloor\sqrt{n}\rfloor+1}}^n 2^{-\frac{kn}{k+n}}$$
For $k \in [\lfloor\sqrt{n}\rfloor + 1,n] $ we have $$ \frac{kn}{k+n} > \frac{\sqrt{n} n}{n+ n}= \frac{\sqrt{n}}{2}$$ $$ 2^{-\frac{kn}{k+n}} < 2^{-\frac{\sqrt{n}}{2}}$$ We have then $$ \sum_{k={\lfloor\sqrt{n}\rfloor+1}}^n 2^{-\frac{kn}{k+n}} < (n-\lfloor\sqrt{n}\rfloor) 2^{-\frac{\sqrt{n}}{2}} < n2^{-\frac{\sqrt{n}}{2}} \rightarrow 0$$ so $$ \lim_{n\rightarrow\infty} \sum_{k={\lfloor\sqrt{n}\rfloor+1}}^n 2^{-\frac{kn}{k+n}} = 0$$
For $k \in [0,\lfloor\sqrt{n}\rfloor]$ we have $$ \frac{kn}{\sqrt{n}+n}\le \frac{kn}{k+n} \le k $$ $$ 2^{-k} \le 2^{-\frac{kn}{k+n}} \le 2^{-\frac{kn}{\sqrt{n}+n}} $$ $$ \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-k} \le \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} \le \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{\sqrt{n}+n}} $$ $$ \frac{1-2^{-\lfloor\sqrt{n}\rfloor-1}}{1-2^{-1}} \le \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} \le \frac{1-2^{-\frac{(\lfloor\sqrt{n}\rfloor+1)n}{\sqrt{n}+n}}}{1-2^{-\frac{n}{\sqrt{n}+n}}} $$ We have $$ \lim_{n\rightarrow\infty} \frac{1-2^{-\lfloor\sqrt{n}\rfloor-1}}{1-2^{-1}} = \lim_{n\rightarrow\infty} \frac{1-2^{-\frac{(\lfloor\sqrt{n}\rfloor+1)n}{\sqrt{n}+n}}}{1-2^{-\frac{n}{\sqrt{n}+n}}} = 2 $$ which means that also $$ \lim_{n\rightarrow\infty} \sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} = 2 $$
Altogether we have $$ \lim_{n\rightarrow\infty} \sum_{k=0}^n 2^{-\frac{kn}{k+n}} = \lim_{n\rightarrow\infty} \Big(\sum_{k=0}^{\lfloor\sqrt{n}\rfloor} 2^{-\frac{kn}{k+n}} + \sum_{k={\lfloor\sqrt{n}\rfloor+1}}^n 2^{-\frac{kn}{k+n}}\Big) = 2 + 0 =2$$