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PostPosted: Thu Jul 07, 2016 1:42 pm 

Joined: Mon Nov 09, 2015 12:29 am
Posts: 36
Show that \(\displaystyle\sum_{n\geq1}\frac{(-1)^{n-1}}{n(n+1)\cdots(n+k)}=\frac{2^k}{k!}\left(\ln2-\sum_{i=1}^{k}\frac{(1/2)^i}{i}\right)\) where \(k\) is a non negative integer and when \(k=0\) the second sum is to be interpreted as the sum over the empty set, i.e. \(=0\).


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PostPosted: Thu Jul 07, 2016 1:42 pm 

Joined: Mon Nov 09, 2015 11:52 am
Posts: 77
Location: Limassol/Pyla Cyprus
Let us begin by recalling that the Beta function satisfies \[ B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt = \frac{(x-1)!(y-1)!}{(x+y-1)!}\] whenever \(x,y\) are positive integers.

In our case we have \[ \sum_{n \geqslant 1} \frac{(-1)^{n-1}}{n(n+1) \cdots (n+k)} = \frac{1}{k!} \sum_{n\geqslant 1} (-1)^{n-1}B(k+1,n) = \frac{1}{k!}\sum_{n\geqslant 1} (-1)^{n-1}\int_0^1 t^k(1-t)^{n-1} \,dt.\] Since the convergence is absolute for \(k \geqslant 1\) we can use Fubini's theorem to interchange the order of summation and integration to get that the sum is equal to \[\frac{1}{k!} \int_0^1 t^k \sum_{n\geqslant 1} (t-1)^n \, dt = \frac{1}{k!} \int_0^1 \frac{t^k}{2-k} \, dt.\] Dividing \(t^k\) by \(t-2\) we have a quotient of \(t^{k-1} + 2t^{k-2} + \cdots + 2^{k-2}t + 2^{k-1}\) and a remainder of \(2^k\). Therefore the sum is equal to \[ \frac{2^k}{k!} \int_0^1 \frac{1}{2-t} \, dt - \frac{1}{k!} \int_0^1 \sum_{i=1}^k t^{i-1}2^{k-i} \, dt\] which is equal to the required result.

The case \(k=0\) cannot be justified by Fubini. We can however justify it using the dominated convergence theorem on the sequence \(f_n(t)\) defined by \[ f_n(t) = \sum_{r=1}^n (t-1)^{r-1} = \frac{1-(t-1)^n}{2-t}\] which is dominated by the constant function \(g(t) = 2\) on \([0,1]\). In a more elementary way we can observe that \[ 1 - \frac{1}{2} + \frac{1}{3} - \cdots - \frac{1}{2n} = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}\] and then use the fact that the harmonic series is asymptotically equal to \(\log{n}\) to get the result.


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PostPosted: Thu Jul 07, 2016 1:44 pm 

Joined: Mon Nov 09, 2015 12:29 am
Posts: 36
Nice solution. This problem is I generalization of problem 158 (http://people.missouristate.edu/lesreid/Adv158.html) I sent to Missouri State University Problem Corner. Here is my approach:

Using the notation \(\Gamma(k+1)=k!\) for \(k\in\mathbb{N}\cup\{0\}\), where \(\Gamma\) is the Gamma function, we will show that

$$\sum_{k\geq1}(-1)^{k-1}\frac{1}{k\cdot(k+1)\cdots(k+m)}=\frac{2^m}{\Gamma(m+1)}\left(\ln2-\sum_{k=1}^{m}\frac{\left(1/2\right)^k}{k}\right),\hspace{10ex}m\in\mathbb{N}.$$

Using the identity \(\Gamma(k+1)=k\Gamma(k),\quad k>0\), by a direct calculation we see that

\begin{equation}\label{1}\frac{\Gamma(k)}{\Gamma(k+m+1)}=\frac{1}{m}\left(\frac{\Gamma(k)}{\Gamma(k+m)}-\frac{\Gamma(k+1)}{\Gamma(k+m+1)}\right),\hspace{10ex}m\in\mathbb{N}.\end{equation}

Now setting \(A_{m,n}:=\sum_{k=1}^{n}(-1)^{k-1}\frac{1}{k\cdot(k+1)\cdots(k+m)},\quad m\in\mathbb{N}\cup\{0\}\), on account of \((1)\) we have:

\(\begin{align}A_{m,n}&=\sum_{k=1}^{n}(-1)^{k-1}\frac{\Gamma(k)}{\Gamma(k+m+1)}\notag \\ &=\frac{1}{m}\sum_{k=1}^{n}(-1)^{k-1}\left(\frac{\Gamma(k)}{\Gamma(k+m)}-\frac{\Gamma(k+1)}{\Gamma(k+m+1)}\right)\notag \\&=\frac{1}{m}\left(\frac{1}{\Gamma(m+1)}+(-1)^n\frac{\Gamma(n+1)}{\Gamma(n+m+1)}+2\sum_{k=2}^{n}(-1)^{k-1}\frac{\Gamma(k)}{\Gamma(k+m)}\right)\notag\\&=\frac{1}{m}\left(\frac{1}{\Gamma(m+1)}+(-1)^n\frac{\Gamma(n+1)}{\Gamma(n+m+1)}+2\left(A_{m-1,n}-\frac{1}{\Gamma(m+1)}\right)\right),\notag\end{align}\)

so \(\displaystyle A_{m,n}=\frac{1}{m}\left(-\frac{1}{\Gamma(m+1)}+(-1)^{n}\frac{\Gamma(n+1)}{\Gamma(n+m+1)}+2A_{m-1,n}\right),\) and, since for \(m\in\mathbb{N}\cup\{0\}\) by Dirichlet's criterion \(\displaystyle\sum_{k\geq1}(-1)^{k-1}\frac{\Gamma(k)}{\Gamma(k+m+1)}\) converges and for \(m\in\mathbb{N}\) we have \(\displaystyle(-1)^{n}\frac{\Gamma(n+1)}{\Gamma(n+m+1)}\to0\), setting \(\displaystyle A_m:=\sum_{k\geq1}(-1)^{k-1}\frac{\Gamma(k)}{\Gamma(k+m+1)}\) and letting \(n\to+\infty\) we have

\(\begin{equation}\label{2}\displaystyle A_m=-\frac{1}{m\Gamma(m+1)}+\frac{2}{m}A_{m-1},\hspace{3ex}m\in\mathbb{N}.\end{equation}\)

Since \(A_{0}=\ln2\), with a simple inductive argument, \((2)\) yields

\(\displaystyle A_m=\frac{2^m}{\Gamma(m+1)}\left(\ln2-\sum_{k=1}^{m}\frac{\left(1/2\right)^k}{k}\right),\hspace{10ex}m\in\mathbb{N}\cup\{0\}.\)


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