Limit of pseudo-series 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?
 A: First, it is straightforward to see that 
$$\sum_{k=0}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}\ge \sum_{k=0}^\infty\left(\frac{1}{2^k}\right)=2$$

Next, we write for a fixed number $0<K<2n$
$$\begin{align}
\sum_{k=0}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}} &=\sum_{k=0}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}\\\\
&=\sum_{k=0}^K\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}+\sum_{k=K+1}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}\tag1
\end{align}$$

Given a number $\epsilon>0$, we choose $K$ large enough so that both 
$$\left(\frac12\right)^K<\epsilon/2$$
and 
$$\frac{\left(\frac12\right)^{K/3}}{2^{1/3}-1}<\epsilon/2$$  

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

Putting it all together shows that 
$$\lim_{n\to\infty}\sum_{k=0}^{2n}\left(\frac{1}{2^k}\right)^{\frac{1}{1+k/n}}=2$$
A: 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.
