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!

  • $\begingroup$ can you show us your working? $\endgroup$ – danimal Apr 17 '15 at 11:06
  • $\begingroup$ @Henry which book is this? $\endgroup$ – Kugelblitz Jun 17 '17 at 4:00

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,

$\text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}}$

$\implies \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} \ \left[\text{since} \dbinom{n}{r}= \dfrac{n}{r} \times \dbinom{n-1}{r-1} \right]$


$$ \text{S}(n) = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}} = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{n-r+1}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{n-r}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} $$

$ \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.


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

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.

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  • 1
    $\begingroup$ Wow! Ingenious and rigorous as always :) But can you please tell what led you to think this way? $\endgroup$ – Henry Durham Apr 17 '15 at 13:06
  • $\begingroup$ @Samurai Your question reminded me of Beta functions which I have used to create this solution. Although understanding the solution doesn't need the understanding of Beta functions, but if you look carefully at Lemma 2, it is the definition of Beta function, for which I've provided an elementary proof. The rest followed from rigorous brainstorming :) $\endgroup$ – MathGod Apr 17 '15 at 13:11
  • $\begingroup$ @MathGod Thanks a lot :D $\endgroup$ – Henry Durham Apr 17 '15 at 13:14
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    $\begingroup$ @MathGod Would you mind explaining how you got: $$\int_{0}^1 (n+1) \sum_{r=0}^n (1-x)^n \left(\frac{x}{1-x}\right)\mathrm{d}x$$ after $$\int_{0}^1 (n+1) (1-x)^n \left(\dfrac{\left(\frac{x}{1-x}\right)^{n+1}-1}{\frac{x}{1-x}-1}\right) \mathrm{d}x$$ It is not so clear to me. Thank you. $\endgroup$ – Tolaso Apr 18 '15 at 4:33
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    $\begingroup$ @Tolaso Thanks for pointing that out. It was a minor typo and I've now corrected it. The expression is sum of G.P. $\endgroup$ – MathGod Apr 18 '15 at 6:05

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.

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  • $\begingroup$ Pardon me, but I don't know what's dominated convergence theorem. $\endgroup$ – Henry Durham Apr 17 '15 at 11:33
  • $\begingroup$ @Samurai: en.wikipedia.org/wiki/Dominated_convergence_theorem $\endgroup$ – Jack D'Aurizio Apr 17 '15 at 11:35
  • $\begingroup$ That's sweet... I guess, without knowing the properties of $H_n$ you couldn't do much... But how did you get the first equality in $(2)$? $\endgroup$ – Igor Deruga Apr 17 '15 at 11:43
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    $\begingroup$ @Igor: $t=(1+t)-1$ and $1+z+\ldots+z^n = \frac{z^{n+1}-1}{z-1}$. $\endgroup$ – Jack D'Aurizio Apr 17 '15 at 11:48

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 $2$.

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