Evaluating $\lim_{n \to \infty} \int_{0}^{1} n(2^x-1)^n dx$ I tried to calculate this dominant convergence theorem to solve this limit.
Evaluate the following limit for $x=0$ and for $x>0$:
$$\lim_{n \to \infty} \int_{0}^{1} n(2^x-1)^n dx$$
and I got the limit value : $\frac{1}{2\ln{2}}$
did I do the right thing?
 A: The following result has appeared on MSE many times: If $f$ is continuous on $[0,1],$ then
$$\tag 1 n\int_0^1f(u)u^n\,du \to f(1).$$
The dominated convergence theorem gives a quick proof. See For $f$ continuous, show $\lim_{n\to\infty} n\int_0^1 f(x)x^n\,dx = f(1).$
Assuming this, let $u=2^x-1.$ Then our integral becomes
$$\frac{1}{\ln 2}\int_0^1 \frac{u^n}{u+1}\,du.$$
By $(1),$ the last expression $\to \dfrac{1}{\ln 2}\cdot \dfrac{1}{2}.$
A: Let $$a_n=\int_{0}^1 (2^x-1)^n\,dx$$
Then $a_0=1,$ and
$$\begin{align}a_{n+1} &=\left(\int_0^1 (2^x-1)^n2^x\,dx \right)-a_n\\
&=-a_n +\frac{1}{\log 2}\int_1^2 (u-1)^n\,du\quad \quad (u=2^x)\\
&=-a_n+\frac{1}{(n+1)\log 2}
\end{align}$$
Let $b_n=na_n.$
It takes some work to prove that $b_n$ converges.
Show by induction that $$a_n=(-1)^n\left(1-\frac1{\log2}\sum_{k=1}^n\frac{(-1)^{k-1}}k\right)$$
We know $s_n=\sum_{k=1}^n\frac{(-1)^{k-1}}k\to\log 2,$
Because $s_n$ is a decreasing and alternating series, we now that:
$$|s_{n+1}-\log 2|<|s_{n}-\log 2|$$
So $s_n+r_n\frac{(-1)^n}{n+1}=\log 2$ for some $r_n\in(1/2,1).$   Similarly, $s_n+q_n\frac{(-1)^n}n=\log 2$ for some $q_n\in(0,1/2).$ So $\frac{q_n}{n}=\frac{r_n}{n+1}$ or $$r_n-q_n=\frac{q_n}{n}$$ and thus $r_n-q_n\to 0,$ and thus $r_n,q_n\to \frac12.$
This shows the limit exists, since $$b_n=\frac{n}{\log2}(-1)^n(\log2-s_n)=\frac{q_n}{\log 2}.$$

More generally,

Theorem: if $c_n$ is positive, decreasing, $c_n\to 0,$ and $\frac{c_n}{c_{n+1}}\to 1.$ Let  $s_n=\sum_{k=1}^n (-1)^{k-1}c_k,$ and $L=\lim s_n.$ Then $$\lim \frac{|s_n-L|}{c_n}=\frac{1}{2}.$$

Facts:
We assume we know that:

*

*For all $n,$ $$s_{2n}<L<s_{2n-1}.$$

*For all $n,$ $$|s_{n+1}-L|<|s_n-L|.$$
Proof of theorem: As before, $$L=s_n+r_n(-1)^nc_{n+1}=s_n+q_n(-1)^nc_n$$
for some $0<q_n<\frac12<r_n<1.$ The existence of these is ensures because of facts 1. and 2. above - specifically, 1. ensures $r_n,q_n\in(0,1)$ and 2. ensures that $q_n<1/2<r_n.$ (In face $q_n=1-r_{n-1}.$)
Then: $r_nc_{n+1}=q_nc_n$ or: $$r_n-q_n=\left(\frac{c_n}{c_{n+1}}-1\right)q_n\to 0.$$
So $q_n,r_n\to\frac12.$ But:$$\frac{|s_n-L|}{c_n}=q_n.$$
