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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?

kimchi lover
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popmaria
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    It doesn’t make sense to say “evaluate this limit for $x=0$ and for $x>0.$” The variable $x$ doesn’t take a single value in the formula, but it runs through all haves in$[0,1].$ – Thomas Andrews Jun 16 '21 at 14:11
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    You should show us your calculations – Thomas Jun 16 '21 at 14:27

2 Answers2

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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 $\to \dfrac{1}{\ln 2}\cdot \dfrac{1}{2}.$

zhw.
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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:

  1. For all $n,$ $$s_{2n}<L<s_{2n-1}.$$
  2. 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:$$\frac{|s_n-L|}{c_n}=q_n.$$

Thomas Andrews
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