Evaluate $\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$ 
Evaluate 
  $$\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$  

For this integral, I have tried using integration by parts and then evaluating the limit, but I don't think the integral inside converges. However, the limit does exist and the answer given in my book is $2$.   
Any help will be appreciated.
Thanks in advance!
 A: It is worth to notice that:
$$\int_{0}^{1}\frac{1-(1-t)^n}{t}\,dt = H_n\tag{1}$$
while:
$$\int_{0}^{1}\frac{(1+t)^n-1}{t}\,dt=\int_{0}^{1}\sum_{k=0}^{n-1}(1+t)^k\,dt =\sum_{k=1}^{n}\frac{2^k-1}{k}\tag{2}$$
so:
$$J_n=\int_{0}^{1}\frac{(1+t)^n-(1-t)^n}{t}\,dt = \sum_{k=1}^{n}\frac{2^k}{k}.\tag{3}$$
The last line also gives:
$$ \frac{n\, J_n}{2^n} = \sum_{k=1}^{n}\frac{2n}{n+1-k}\cdot 2^{-k} \tag{4}$$
and when $n$ approaches $+\infty$, by the dominated convergence theorem the RHS of $(4)$ approaches:
$$\sum_{k=1}^{+\infty}2\cdot 2^{-k} = \color{red}{2} \tag{5}$$
as wanted.
A: First, consider the following two Lemmas,  
Lemma $1$: $$\lim_{n \to \infty} \sum_{r=0}^n \left(\dfrac{1}{\displaystyle\binom{n}{r}}\right) =2$$    
Proof : First of all, note that the limit exists, since, if we let
$$\text{S}(n)=\displaystyle \sum_{r=0}^n \left(\dfrac{1}{\dbinom{n}{r}} \right)$$  
then $\text{S}(n+1)<\text{S}(n)$ for $n \geq 4$. Now,  
$\text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}}$  
$\implies \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} \ \left[\text{since} \dbinom{n}{r}= \dfrac{n}{r} \times \dbinom{n-1}{r-1} \right]$  
Also,  
$$ \text{S}(n) = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}} = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{n-r+1}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{n-r}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} $$  
$ \left[ \text{since} \ \displaystyle \sum_{r=a}^b f(r) = \displaystyle \sum_{r=a}^b f(a+b-r) \ \text{and} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]$  
Thus, we have,  
$$\begin{cases} \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}}\\ \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} \end{cases}$$  
Adding the above two expressions, we get, 
$ 2\text{S}(n) = 2 + \displaystyle \sum_{r=1}^n \left( \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} + \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} \right) $  
$= 2 + \dfrac{n+1}{n} \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n-1}{r-1}} $  
$= 2 + \dfrac{n+1}{n} \times \text{S}(n-1)$  
Since $ n \to \infty $, we have $\text{S}(n) = \text{S}(n-1) = \text{S}$ (say)  
$ \implies 2\text{S} = \left(\dfrac{n+1}{n}\right) \times \text{S} +2 $  
$ \implies \text{S} = \dfrac{2n}{n-1} = 2$ [since $n \to \infty $]  
Lemma $2$ :  $$\int_{0}^1 x^r (1-x)^{n-r} \mathrm{d}t = \dfrac{1}{(n+1)}\times \dfrac{1}{\dbinom{n}{r}}$$  
Proof : Consider the $\text{R.H.S.}$,  
$\text{I} = \displaystyle\int_{0}^1 x^r (1-x)^{n-r} \mathrm{d}t$   
Let $x = \sin^2 \theta$  
$\implies \text{I} = \displaystyle\int_{0}^{\frac{\pi}{2}} 2 \sin^{2r+1} \theta \cos^{2n-2r} \theta \ \mathrm{d}\theta $   
Now, using Walli's Formula (or reduction formula) for the above integral, we have,   
$\text{I} = \dfrac{1}{(n+1)}\times \dfrac{1}{\dbinom{n}{r}} $ 
This proves our Lemmas.
Now, 
$$ \text{J} = \lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t $$  
Since it is an even function in $t$, we have,  
$$ \text{J} = \frac{1}{2} \times \lim_{n \to \infty} \int_{-1}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t $$  
Let $t = 2x-1$  
$\implies \text{J} = \displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \left(\dfrac{x^{n+1}-(1-x)^{n+1}}{2x-1}\right) \mathrm{d}x$  
$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) (1-x)^n \left(\dfrac{\left(\frac{x}{1-x}\right)^{n+1}-1}{\frac{x}{1-x}-1}\right) \mathrm{d}x$  
$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \sum_{r=0}^n (1-x)^n \left(\frac{x}{1-x}\right)^{r} \mathrm{d}x$  
$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n (n+1) \int_{0}^1 x^r(1-x)^{n-r} \mathrm{d}x$  
$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n \dfrac{1}{\dbinom{n}{r}}$ (Using Lemma 2) 
$=\boxed{2}$ (Using Lemma 1). 
Side Note : Another way to prove Lemma 1 is to use sandwich theorem.
A: We need to address area around $0$ separately from area around $1$. Pick $\epsilon\in(0,1)$. Then on $[0,\epsilon]$
$$\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\le (n+1)(1+\epsilon)^{n}+(n+1)(1-\epsilon)^{n}$$(you can see this if you multiply both sides by $t$ and compare derivatives) and hence
$$\lim_{n\to\infty}\int_0^{\epsilon}\frac {n+1}{2^{n+1}}\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\,dt=0$$
On the other hand
$$\lim_{n\to\infty}\int_\epsilon^1\frac {n+1}{2^{n+1}}\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\,dt=\lim_{n\to\infty}\int_{\epsilon}^1\frac 2 t\frac {n+1}{n+2} d\Big(\big(\frac {1+t}{2}\big)^{n+2}+\big(\frac {1-t} 2\big)^{n+2}\Big)=\int_{\epsilon}^1\frac 2 t dI(t=1)=2$$
since $\frac 2 t$ is bounded on $[\epsilon,1]$. Hence the original limit is $2$.
