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It can be shown that

$$ n \int_0^1 \frac{x^n - (1-x)^n}{2x-1} \mathrm dx = \sum_{k=0}^{n-1} {n-1 \choose k}^{-1}$$

(For instance see my answer here.)

It can also be shown that $$\lim_{n \to \infty} \ \sum_{k=0}^{n-1} {n-1 \choose k}^{-1} = 2$$

(For instance see Qiaochu's answer here.)

Combining those two shows that

$$ \lim_{n \to \infty} \ n \int_0^1 \frac{x^n - (1-x)^n}{2x-1} \mathrm dx = 2$$

Is there a different, (preferably analytic) proof of this fact? Please do feel free to add a proof which is not analytic.

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What do u mean by an analytic proof –  anonymous Oct 11 '10 at 18:20
    
@Chandru: Using theorems in real/complex analysis etc. For instance, without the n factor, the limit is zero, and that follows from the dominated convergence theorem. –  Aryabhata Oct 11 '10 at 18:22
    
OK! Understood –  anonymous Oct 11 '10 at 18:25
    
Why did you change your name? :) –  AD. Jun 29 '11 at 20:49
    
@AD.: I was thinking of changing it and the comments here helped: math.stackexchange.com/questions/38473/… :-) –  Aryabhata Jun 29 '11 at 20:52
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3 Answers

up vote 13 down vote accepted

Let $x=(1+s)/2$, so that the expression becomes $$ \frac{n}{2^{n+1}} \int_{-1}^1 \frac{(1+s)^n-(1-s)^n}{s} ds = \frac{n}{2^{n}} \int_{0}^1 \frac{(1+s)^n-(1-s)^n}{s} ds. $$ (The integrand is an even function.) Fix some $c$ between 0 and 1, say $c=1-\epsilon$. Then the integral from 0 to $c$ will be small in comparison to $2^n$, since the integrand is bounded by a constant times $(1+c)^n$, so as far as the limit is concerned it is enough to look at the integral from $c$ to 1, and in that integral we can neglect the term $(1-s)^n$ since it will also contribute something much smaller than $2^n$. The surviving contribution therefore comes from $$ \int_c^1 \frac{(1+s)^n}{s} ds, $$ which lies between $$ \int_c^1 \frac{(1+s)^n}{1} ds $$ and $$ \int_c^1 \frac{(1+s)^n}{c} ds, $$ that is, $$ \frac{2^{n+1} - (1+c)^{n+1}}{n+1} < \int_c^1 \frac{(1+s)^n}{s} ds < \frac{1}{c} \frac{2^{n+1} - (1+c)^{n+1}}{n+1} .$$ Multiplying by $n/2^n$ and letting $n\to\infty$ shows that the liminf is at least 2 and the limsup is at most $2/c$. But since this holds for any $c$ between 0 and 1, it follows that liminf=limsup=2, hence the limit is 2.

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Thanks for this! Though, one thing that seems to be missing is the proof that the limit exists, which I suppose won't be too hard to show (it is not exactly sandwich theorem, but perhaps a 2D variant of it). –  Aryabhata Oct 12 '10 at 16:08
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@Moron: Doesn't the last step show that the liminf is at least 2, and the limsup is at most 2/c, for all c<1? Hence liminf=limsup=2. –  Hans Lundmark Oct 12 '10 at 17:13
    
Yes that proves it :-), but the last paragraph talks about 'the limit' before doing the squeeze between 2 and 2/c as c->1. I suggest you change the wording to talk about liminf and limsup instead of 'the limit', to make that clear. Sorry, i guess it is just a nitpick. –  Aryabhata Oct 12 '10 at 17:25
    
btw, the only reason for the request is that it was not obvious to me (until you pointed it out :-)) and I am guessing spelling it out might help others in the future. –  Aryabhata Oct 12 '10 at 17:48
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@Moron: Done! Thanks for pointing it out. Being a nitpick is a good thing in mathematical analysis. :) –  Hans Lundmark Oct 13 '10 at 6:43
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Here's another idea.

Let $P_n(x)$ be the polynomial defined by $$P_n(x) = \frac{x^n – (1-x)^n}{2x-1},$$ and note that $P_n(x)=(1-x)P_{n-1}(x) + x^{n-1} \qquad (1), $ where $P_1(x)=1.$

Define $I_n = \int_0^{1} P_n(x) dx.$ Integrating (1) between 0 and 1 we obtain

$$I_n = \int_0^{1} (1-x)P_{n-1}(x) dx + \frac{1}{n},$$

and using the symmetry of $P_n(x)$ about 1/2 we note that $\int_0^{1} (1-x)P_{n-1}(x) dx = \frac{1}{2}I_{n-1}.$ Hence we obtain the recurrence relation for $n > 1$ $$I_n = \frac{1}{2}I_{n-1} + \frac{1}{n}, \textrm{ where } I_1=1. \qquad (2)$$

From which it follows that if $\lim_{n \rightarrow \infty} n I_n$ exists then it must equal 2.

To show that the limit exists we can solve (2) to obtain

$$I_n = \sum_{k=1}^n \frac {1}{k2^{n-k}},$$

and straightforward algebra shows that $(n-1)I_{n-1} – nI_n > 0$ for sufficiently large $n$ (in fact, $n>6$), and since $nI_n$ is bounded below by 0 the result follows.

Just for completeness:

$$(n-1)I_{n-1} – nI_n = \left\lbrace \sum_{k=1}^{n-2} \frac{k}{2^k(n-k)(n-k-1)} \right\rbrace - \frac{n}{2^{n-1}} \quad \textrm{ for } n>2.$$

Comparing the first term in the summation with the negative term we have

$$ \frac{1}{2(n-1)(n-2)} > \frac{n}{2^{n-1}} \quad \textrm{ for } n \ge 13.$$

Hence $\lbrace nI_n \rbrace$ eventually forms a strictly decreasing sequence.

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Nice. Thanks... –  Aryabhata Oct 13 '10 at 1:28
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Here's a sketch of an elementary analytic demonstration.

(1) The integrals all exceed $2/n$ when $x$ exceeds 1: subtract $2/n = \int_0^1{(t^{n-1} + (1-t)^{n-1}) \ dt}$ to obtain a positive integrand (using elementary inequalities).

(2) The integrals are bounded by a sequence decreasing to the desired limit. By looking at the Taylor series of the log of the integrand near 1, for example, and assuming $n \ge 3$, you can easily establish an upper bound of $\exp((n-2)(x-1))$ in the interval $(1/2, 1]$, with a similar bound in the interval $[0, 1/2)$ by symmetry. The estimate (integrating over one of those two intervals) equals $(1 - \exp(2-n)) / (n-2)$ which, when doubled and multiplied by $n$, converges to $2$.

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Interesting. Is it possible for you to add a little more detail for part 2? It would make it easier for anyone (not just me) who reads your answer. –  Aryabhata Oct 11 '10 at 20:26
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