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 Post subject: A class of alternate infinite seriesPosted: 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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 Post subject: Re: A class of alternate infinite seriesPosted: 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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 Post subject: Re: A class of alternate infinite seriesPosted: 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

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

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

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

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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