$\lim_{n \to \infty} \frac{2^n}{n} \left[ 1 - \sum_{k = 1}^{n-1} \frac{(1 - \lambda 2^{-n})^{2^k}}{2^k} \right]$

Question

Find the limit \begin{equation*} \lim_{n \to \infty} \frac{2^n}{n} \left[ 1 - \sum_{k = 1}^{n-1} \frac{(1 - \lambda 2^{-n})^{2^k}}{2^k} \right] \end{equation*} where $\lambda > 0$.

My guess is that the limit is equal to $\lambda$.

Answer 1

Yes, it is $\lambda$. At first, if $n$ is such that $\lambda 2^{-n}<1$ (therefore, for all large enough $n$) we have $(1-\lambda 2^{-n})^{2^k}\geqslant 1-\lambda 2^{k-n}$ be Bernoulli inequality that yields $$ 1 - \sum_{k = 1}^{n-1} \frac{(1 - \lambda 2^{-n})^{2^k}}{2^k} \leqslant 1-\sum_{k=1}^{n-1}\frac{1-\lambda 2^{k-n}}{2^k}=\lambda(n-1)2^{-n}+2^{1-n}, $$ therefore the upper limit of your expression does not exceed $\lambda$.

For the lower bound, we use that for $x\in [0,1]$ and positive integer $m$ we have $(1-x)^m\leqslant 1-mx+{m\choose 2}x^2$ (for $x=0$ the equality holds, for $x\in [0,1]$ the derivative of RHS-LHS equals $m((1-x)^{m-1}-(1-(m-1)x))\geqslant 0$ by Bernoulli inequality.) Therefore, $$ 1 - \sum_{k = 1}^{n-1} \frac{(1 - \lambda 2^{-n})^{2^k}}{2^k} \geqslant \lambda(n-1)2^{-n}+2^{1-n}-\sum_{k=1}^{n-1}{2^k\choose 2}\lambda^2 2^{-2n-k}, $$ the last sum is $O(2^{-n})$, thus the lower limit of your expression is at least $\lambda$.

Answer 2

We have $$(1 - \lambda 2^{-n})^{2^k} = e ^ {2^k \log(1 - \lambda 2^{-n})} = e ^ {2^k ( - \lambda 2^{-n} + O_\lambda(4^{-n}))} = e^{-\lambda 2^{k-n} + O_\lambda(2^{-n})} = (1 + O_\lambda(2^{-n})) e^{-\lambda 2^{k-n}} = e^{-\lambda 2^{k-n}} + O_\lambda(2^{-n}) = 1 -\lambda 2^{k-n} + O_\lambda(2^{2k - 2n}) + O_\lambda(2^{-n}).$$ So we have $$\frac1n\left(2^n + O_{\lambda}(1) - \sum_{k=1}^{n-1} 2^{n-k} (1 -\lambda 2^{k-n} + O_\lambda(2^{2k - 2n}))\right) = \frac1n\left(2^n + O_{\lambda}(1) - \sum_{k=1}^{n-1} 2^{n-k} (1 -\lambda 2^{k-n})\right) = \frac1n\left(O_{\lambda}(1) + \sum_{k=1}^{n-1} \lambda\right).$$ And the limit is indeed $\lambda$.