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American Mathematical Monthly problem 11611 essentially asks you to show that $$\lim_n\ n \int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx=2\log(2).\tag1$$

This would follow easily from (2) below, which is true for small values of $n$ according to Maple. But I couldn't prove equation (2) in general, so I found a direct solution for (1) instead.

$$\int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx.\tag2$$

But I'm still curious about (2). What am I missing? How can equation (2) be proven?

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It should be doable by induction, though it may be not the shortest way. –  Andrew May 2 '12 at 6:22
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@UmNyobe Please post a sketch of such a proof. –  Sasha May 2 '12 at 16:15
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Thank you for waiting until after the April 30 deadline of solutions before asking here... –  GEdgar May 2 '12 at 18:51
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An honest question: Byron how did you conjure up formula (2) in the first place? I'm sure there was some context that lead you to it, and I would love to see the reasoning behind it. –  Alex R. May 2 '12 at 21:42
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@Sam Nothing very profound, I'm afraid! I used Maple to calculate the first few values of the LHS, and plugged the numerators into the On-Line Encyclopedia of Integer Sequences. This led to sequence A049281 which mentions the RHS integral in its comments. –  Byron Schmuland May 2 '12 at 21:49
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4 Answers 4

up vote 10 down vote accepted

One can avoid $\psi$ functions and even the expansion of the logarithm (right, for the logarithm, I am kind of cheating, try to find where...). Call $L_n$ and $R_n$ the LHS and the RHS of (2).

  1. Expand the square of the series in $L_n$ and integrate term by term the resulting double series to get $$ L_n=\sum_{i\geqslant n}\sum_{j\geqslant n}\frac1{ij(i+j+1)}. $$ Since $\color{red}{\dfrac{i+1}{j(j+i+1)}=\dfrac1j-\dfrac1{i+j+1}}$, $$ L_n=\sum_{i\geqslant n}\frac1{i(i+1)}\sum_{j\geqslant n}\left(\frac1j-\frac1{i+j+1}\right)=\sum_{i\geqslant n}\frac1{i(i+1)}\sum_{j\geqslant n}\frac1j[i\geqslant j-n]. $$ Exchanging the order of summations and using the identity $\color{red}{\dfrac1{i(i+1)}=\dfrac1i-\dfrac1{i+1}}$, one gets $$ L_n=\sum_{j\geqslant n}\frac1j\frac1{\max(n,j-n)}=\sum_{j=n}^{2n-1}\frac1j\frac1n+\sum_{j\geqslant2n}\frac1{j(j-n)}. $$ Since $\color{red}{\dfrac{n}{j(j-n)}=\dfrac1{j-n}-\dfrac1j}$, $$ L_n=\frac1n\sum_{j=n}^{2n-1}\frac1j+\frac1n\sum_{j\geqslant2n}\left(\frac1{j-n}-\frac1j\right)=\frac2n\sum_{j=n}^{2n-1}\frac1j. $$
  2. Use the change of variables $x=t^2$ and an integration by parts to get $$ nR_n=2\int_0^1\frac{1+t^{2n-1}}{1+t}\mathrm dt. $$
  3. Compute the differences $$ (n+1)L_{n+1}-nL_n=2\left(\frac1{2n+1}-\frac1{2n}\right), $$ and $$ (n+1)R_{n+1}-nR_n=2\int_0^1\frac{t^{2n+1}-t^{2n-1}}{1+t}\mathrm dt=2\int_0^1(t^{2n}-t^{2n-1})\mathrm dt=2\left(\frac1{2n+1}-\frac1{2n}\right). $$
  4. Compute the initial terms $L_1=R_1=2$ and conclude that $nL_n=nR_n$ for every $n\geqslant1$.
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Thanks for your elegant solution. I have so many wonderful solutions now, it will be hard to choose which one to accept. –  Byron Schmuland May 2 '12 at 18:15
    
It seems this one is based on Abel like transformations obtained three times through the key identity $\frac{a}{b(b+a)}=\frac1b-\frac1{b+a}$ and nothing else, so to speak. There might even be a loop somewhere which would allow to shorten this still more... –  Did May 2 '12 at 18:20
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@Didier +1! Really like your working out of $L_n$. –  Sasha May 2 '12 at 18:33
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Let $\mathcal{R}_n$ denote the integral on the right-hand-side of eq. (2): $$ \mathcal{R}_n = \int_0^1 2 x^{n-1} \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x $$ Consider $$ \begin{eqnarray} (n+1) \mathcal{R}_{n+1} - n \mathcal{R}_n &=& \int_0^1 2 \left( (n+1) x^{n} - n x^{n-1} \right) \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x \\ &=& \int_0^1 2 \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d} \left( -x^n \left(1-x\right)\right) \\ &=& \int_0^1 \left(\sqrt{x}-1\right)x^{n-1} \mathrm{d} x = \frac{2}{2n+1} - \frac{1}{n} \end{eqnarray} $$ Therefore: $$ n \mathcal{R}_n = \mathcal{R}_1 + \sum_{m=1}^{n-1} \left( \frac{2}{2m+1} - \frac{1}{m} \right) = \psi\left(n+\frac{1}{2}\right) - \psi(n) + 2 \left(\log(2)-1\right) + \mathcal{R}_1 $$ Integral $\mathcal{R}_1$ can be easily integrated by parts: $$ \mathcal{R}_1 = 2 \int_0^1 \log\left(1+\frac{1}{\sqrt{x}}\right)\mathrm{d} x\stackrel{\text{by parts}}{=} \left. 2\left( x \log\left( 1 + \frac{1}{\sqrt{x}}\right) + \sqrt{x} - \log\left(1+\sqrt{x}\right) \right) \right|_{x \downarrow 0}^{x = 1} = 2 $$ Thus $$ n \mathcal{R}_n = 2 \log(2) + \psi\left(n+\frac{1}{2}\right) - \psi(n) $$

Similarly, denoting $\mathcal{L}_n = \int_0^1 f_n(x)^2 \mathrm{d} x$ the integral on the left-hand-side of eq. (2): $$ \begin{eqnarray} n \left(\mathcal{L}_{n+1} - \mathcal{L}_n\right) &=& \int_0^1 \left( n f_{n+1}(x)^2 - n \left( \frac{x^n}{n} + f_{n+1}(x) \right)^2 \right)\mathrm{d} x \\ &=& \int_0^1 \left( -2 x^n f_{n+1}(x) - \frac{x^{2n}}{n}\right)\mathrm{d} x \\ &=& \color\maroon{2 \int_0^1 x^n \log(1-x) \mathrm{d} x} + {\color\blue{2 \int_0^1 x^n \sum_{k=1}^{n} \frac{x^k}{k} \mathrm{d} x}} - \frac{1}{n(2n+1)} \\ &=& \color\maroon{-\frac{2}{n+1} H_{n+1}} + 2 \sum_{k=1}^n \frac{1}{k(k+n+1)} - \frac{1}{n(2n+1)} \\ &=& \frac{2}{n+1} \left( \psi(n+1) - \psi(2+2n) \right) - \frac{1}{n(2n+1)} \\ &=& \frac{1}{n (n+1)} \left( \psi(n+1)-\psi\left(n+\frac{3}{2}\right)-2 \log (2) \right) -\frac{1}{n^2 (2 n+1)} \end{eqnarray} $$ where $f_n(x) = \sum_{k=n}^\infty \frac{x^k}{k}$. Now since $\mathcal{L}_1 = \int_0^1 \log^2(1-x)\mathrm{d} x = 2$, and since $\mathcal{R}_{n+1} - \mathcal{R}_n$ equals to $\mathcal{L}_{n+1} -\mathcal{L}_n$, we establish that $\mathcal{L}_n = \mathcal{R}_n$: $$ \mathcal{R}_{n+1} - \mathcal{R}_n = 2\log(2) \left( \frac{1}{n+1}- \frac{1}{n} \right) + \frac{1}{n+1} \left( \psi\left(n+\frac{3}{2}\right) - \psi (n+1)\right) - \frac{1}{n} \left( \psi\left(n+\frac{1}{2}\right) - \psi (n)\right) \stackrel{\color\maroon{\text{use recurrence equation for } \psi}}{=} \mathcal{L}_{n+1} - \mathcal{L}_n $$

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This will take me some time to digest, but I think it is exactly what I was looking for. Thanks! –  Byron Schmuland May 2 '12 at 16:01
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@ByronSchmuland I think I have just simplified the last step of the proof somewhat, eliminating reliance on any CAS. –  Sasha May 2 '12 at 16:14
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Here's a way to do it by brute force. First write $$ \begin{align} \int_0^1\left(\sum_{k = n}^\infty \frac{x^k}{k}\right)^2\,dx & = \sum_{k,m\geq n}\frac{1}{km}\int_0^1x^{k+m}\,dx \\ & = \sum_{k,m \geq n} \frac{1}{km(k+m+1)}. \end{align} $$ Put $r = k+m$, so that $m = r-k$, and transform the sum: $$ \begin{align} \sum_{k,m\geq n} \frac{1}{km(k+m+1)} &= \sum_{r = 2n}^\infty \frac{1}{r+1}\sum_{k = n}^{r-n}\frac{1}{k(r-k)} \\ &= \sum_{r = 2n}^\infty \frac{1}{r(r+1)}\sum_{k=n}^{r-n}\left(\frac{1}{k} + \frac{1}{r-k}\right) \\ & = \sum_{r = 2n}^\infty \frac{2}{r(r+1)}\sum_{k = n}^{r-n}\frac{1}{k} \\ & = 2\sum_{r = 2n}^\infty\left(\frac{1}{r}\sum_{k = n}^{r-n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r-n}\frac{1}{k}\right) \\ & = 2\sum_{r = 2n}^\infty\left(\frac{1}{r}\sum_{k = n}^{r - n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r+1 - n} \frac{1}{k}\right) + 2 \sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)}. \\ \end{align} $$ The first sum in the last line telescopes, so it can be evaluated as $$ 2\sum_{r = 2n}^\infty \left(\frac{1}{r}\sum_{k = n}^{r - n}\frac{1}{k} - \frac{1}{r+1}\sum_{k = n}^{r+1 - n}\frac{1}{k}\right) = \lim_{r \to \infty} \left(\frac{1}{n^2} - \frac{2}{r+1}\sum_{k = n}^{r+1 - n}\frac{1}{k}\right) = \frac{1}{n^2}. $$ Thus we need to prove that $$ \frac{1}{n^2} + 2\sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)} = 2\int_0^1 x^{n-1}\log(1+x^{-1/2})\,dx. $$ Since $$ \begin{align} 2\int_0^1 x^{n-1}\log(1+x^{-1/2})\,dx &= 2\int_0^1 x^{n-1}\log(1 + x^{1/2})\,dx - \int_0^1 x^{n-1}\log{x}\,dx \\ & = 2\int_0^1 x^{n-1}\log(1+x^{1/2})\,dx + \frac{1}{n^2}, \end{align} $$ we need only prove that $$ \sum_{r = 2n}^\infty \frac{1}{(r+1)(r+1 - n)} = \int_0^1 x^{n-1}\log{(1+x^{1/2})}\,dx. $$ This can be done by developing $\log{(1+x^{1/2})}$ in powers of $x^{1/2}$: $$ \begin{align} \int_0^1 x^{n-1}\log{(1+x^{1/2})}\,dx & = \sum_{m = 1}^\infty\frac{(-1)^{m+1}}{m} \int_0^1 x^{n- 1 + m/2}\,dx \\ & = 2\sum_{m = 1}^\infty \frac{(-1)^{m+1}}{m(2n+m)} \\ & = 2 \sum_{\text{$m$ odd}} - 2\sum_{\text{$m$ even}} \frac{1}{m(2n+m)} \\ & = 2 \sum - 4\sum_{\text{$m$ even}} \frac{1}{m(2n+m)} \\ & = 2 \sum_{m = 1}^\infty \frac{1}{m(2n+m)} - \sum_{m = 1}^\infty \frac{1}{m(n+m)} \\ & = 2 \sum_{m = 1}^\infty \frac{1}{m}\left(\frac{1}{2n+m} - \frac{1}{2n+2m}\right) \\ & = \sum_{m = 1}^\infty \frac{1}{(2n+m)(n+m)} \\ & = \sum_{r = 2n}^\infty \frac{1}{(r + 1)( r +1 - n)}, \end{align} $$ which proves the identity. The other methods are obviously more concise, but after working through this I couldn't resist posting it.

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Thanks for all the hard work you put into this answer! –  Byron Schmuland May 2 '12 at 17:59
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I did some symbol calculation on MATHEMATICA, and I found equation (2) might be deduced in such a way:

1)Let $F(n)=2\int_{0}^{1}x^{n-1}\log(1+1/\sqrt{x})dx$

then $F(n)=\frac{1+n\log(4)}{n^2}-\frac{2}{n}\Phi(-1,1,2n+1)$

$=\frac{1+n\log(4)}{n^2}+\frac{1}{n}(\psi(1/2+n)-\psi(1+n))$

where $\Phi(z,s,a)$ is the Lerch transcendent,$\psi(z)$ is the polygamma function.

2)LHS is equal to the double series $G(n)=\sum_{i,j\geq n}\frac{1}{ij(i+j+1)}$

then $G(n)-G(n+1)=\frac{1}{n^2(2n+1)}+2\sum_{i=n+1}^{\infty}\frac{1}{ni(i+n+1)}$

$=\frac{1}{n^2(2n+1)}-2\frac{(\psi(1+n)-\psi(2+2n))}{n(n+1)}$

If $G(n)-G(n+1)-(F(n)-F(n+1))=0$,then induction works.

3)The last step is to prove an equality of polygamma function.I use MATHEMATICA's FullSimplify function, and the result is 0.

This is not a rigorous proof. I hope this will help.

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@y zhao Thanks very much for this intriguing answer! –  Byron Schmuland May 2 '12 at 16:02
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