Limit of pseudo-series

Question

I'm trying to prove that $$\displaystyle\lim_{n\to +\infty}\sum_{k=0}^{2n}2^{-k\frac{n}{n+k}} = \sum_{k=0}^{+\infty}2^{-k} = 2$$ yet I'm a little stuck. I can quite easily prove that $$\displaystyle u_n := \sum_{k=0}^{2n}2^{-k\frac{n}{n+k}}$$ is bounded, but I struggle to do more. Have you got an idea of the proof?

Answer 1

First, it is straightforward to see that

$$\sum_{k=0}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}\ge \sum_{k=0}^\infty\left(\frac{1}{2^k}\right)=2$$

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Next, we write for a fixed number $0<K<2n$

$$\begin{align} \sum_{k=0}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}} &=\sum_{k=0}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}\\\\ &=\sum_{k=0}^K\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}+\sum_{k=K+1}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}\tag1 \end{align}$$

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Given a number $\epsilon>0$, we choose $K$ large enough so that both

$$\left(\frac12\right)^K<\epsilon/2$$

and

$$\frac{\left(\frac12\right)^{K/3}}{2^{1/3}-1}<\epsilon/2$$

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Clearly, we have

$$\lim_{n\to \infty}\sum_{k=0}^K\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}=2-\left(\frac12\right)^K\tag2$$

and

$$\lim_{n\to \infty}\sum_{k=K+1}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}\le \sum_{K+1}^\infty \left(\frac{1}{2^k}\right)^{1/3}=\frac{\left(\frac12\right)^{K/3}}{2^{1/3}-1}\tag3$$

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Putting it all together shows that

$$\lim_{n\to\infty}\sum_{k=0}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}=2$$

Answer 2

You can rewrite $$ \sum_{k=0}^{2n} 2^{-k\frac{1}{1+\frac{k}{n}}} = \sum_{k=0}^\infty f_n(k) $$ with $f_n$ defined for $x\geq 0$ as $$ f_n(x) = 2^{-x\frac{1}{1+\frac{x}{n}}}\mathbb{1}_{[0,2n]}(k)\,. $$

Note that (i) $f_n$ converges pointwise to $f(x) = 2^{-x}$, and that (ii) for all $x$ and $n$, we have $$ 0\leq f_n(x) \leq \frac{1}{2^{x/3}} \stackrel{\rm def}{=} g(x) $$ which is summable. Thus, by the dominated convergence theorem, $$ \lim_{n\to\infty} \sum_{k=0}^\infty f_n(k) = \sum_{k=0}^\infty f(k) = \sum_{k=0}^\infty \frac{1}{2^k} = \boxed{2} $$ as claimed.

Note that the monotone convergence theorem, instead of the DCT, would work as well here.