Summation of reciprocal products

Question

When studying summation of reciprocal products I found some interesting patterns.

$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$

$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$

$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)}=\frac{1}{3\cdot3!}-\frac{1}{3\cdot(N+1)(N+2)(N+3)}$$

The pattern is $$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)(N+2)(N+3)\cdot\cdot\cdot\cdot(N+i)}$$ which is easy to prove by induction. As an easy consequence it follows that $$\lim_{N \to \infty} \sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)(k+3)\cdot\cdot\cdot\cdot(k+i)}\space=\space\frac{1}{i\cdot i!}$$

I did not find this summation of reciprocal products in my mathbooks. Is this a known theorem? Can anyone point me in the right direction for further study?

Answer 1

This identity is a matter of telescoping. Let $q$ be a positive integer.

We obtain \begin{align*} \color{blue}{\sum_{k=1}^N}&\color{blue}{\frac{1}{k(k+1)\cdots(k+q)}}\\ &=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q)}\left((k+q)-k\right)\tag{1.1}\\ &=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q-1)} -\frac{1}{q}\sum_{k=1}^N\frac{1}{(k+1)\cdots(k+q)}\tag{1.2}\\ &=\frac{1}{q}\sum_{k=1}^N\frac{1}{k(k+1)\cdots(k+q-1)}\\ &\qquad\quad-\frac{1}{q}\sum_{k=2}^{N+1}\frac{1}{k(k+1)\cdots(k+q-1)}\tag{1.3}\\ &=\frac{1}{q}\left(\frac{1}{q!}-\frac{1}{(N+1)(N+2)\cdots(N+q)}\right)\tag{1.4}\\ &\,\,\color{blue}{=\frac{1}{q}\left(\frac{1}{q!}-\frac{N!}{(N+q)!}\right)}\tag{1.5}\\ \end{align*} and the claim follows.

Comment:

• In (1.1) we expand numerator and denominator with $q$.

• In (1.2) we split the sums and cancel terms accordingly.

• In (1.3) we shift the index of the right-hand sum and start with $k=2$.

• In (1.4) we do the cancellation thanks to telescoping.

Note: A theory behind it is the calculus of finite differences. In fact this identity is an application of a discrete analogon of the fundamental theorem of calculus. An application of it is given for instance here.

Calculus of finite differences: Let's do the calculation from above a bit more sophisticated.

• In terms of calculus of finite differences we consider the falling factorial operator \begin{align*} x^{\underline{q}}=x(x-1)\cdots(x-q+1) \end{align*}

• which can be expanded to negative integers, written as $-q$ with $q> 0$ which is then defined as \begin{align*} \color{blue}{x^{\underline{-q}}:=\frac{1}{(x+1)(x+2)\cdots(x+q)}}\tag{2.1} \end{align*}

• The forward difference operator $\Delta$ defined as \begin{align*} \left(\Delta F\right)(x):=F(x+1)-F(x) \end{align*} applied to $x^{\underline{-q}}$ gives \begin{align*} \color{blue}{\Delta x^{\underline{-q}}=-qx^{\underline{-q-1}}}\tag{2.2} \end{align*}

• A discrete analogon to the fundamental theorem of calculus is given as \begin{align*} \color{blue}{\sum_{k=0}^{N-1}\left(\Delta F\right)(k)}&\color{blue}{=F(N)-F(0)}\tag{2.3}\\ \int_{0}^xF^{\prime}(t)\,dt&=F(x)-F(0) \end{align*}

We can now perform the calculation from above as \begin{align*} \color{blue}{\sum_{k=1}^N}&\color{blue}{\frac{1}{k(k+1)\cdots(k+q)}} =\sum_{k=1}^{N}(k-1)^{\underline{-\left(q+1\right)}}\tag{$\to\ $2.1}\\ &=\sum_{k=0}^{N-1}k^{\underline{-\left(q+1\right)}}\tag{shift $k$}\\ &=-\frac{1}{q}\sum_{k=0}^{N-1}\left(\Delta k^{\underline{-q}}\right)\tag{$\to\ $2.2}\\ &=-\frac{1}{q}\left(N^{\underline{-q}}-0^{\underline{-q}}\right)\tag{$\to\ $2.3}\\ &=-\frac{1}{q}\left((N+1)(N+2)\cdots(N+q)-1\cdot 2\cdots q\right)\tag{$\to\ $2.1}\\ &\,\,\color{blue}{=\frac{1}{q}\left(\frac{1}{q!}-\frac{N!}{(N+q)!}\right)} \end{align*} and the claim follows again in accordance with the result (1.5).

Note: A generalisation of this identity even more compactly derived is given as formula (1.16) in Differenzen und Summen in Konkrete Analysis by J. Cigler.

Answer 2

I think I've posted this before, but what the heck.

The telescoping occurs not only for the product of consecutive integers but for its reciprocal.

Here's the proofs.

$p_m(x) =\prod_{k=0}^{m-1} (x+k) $.

$\begin{array}\\ p_m(x+1)-p_m(x) &=\prod_{k=0}^{m-1} (x+1+k)-\prod_{k=0}^{m-1} (x+k)\\ &=\prod_{k=1}^{m} (x+k)-\prod_{k=0}^{m-1} (x+k)\\ &=(x+m-x)\prod_{k=1}^{m-1} (x+k)\\ &=m\prod_{k=0}^{m-2} (x+1+k)\\ &=mp_{m-1}(x+1)\\ p_m(x)-p_m(x-1) &=mp_{m-1}(x)\\ \dfrac1{p_m(x)}-\dfrac1{p_m(x+1)} &=\dfrac1{\prod_{k=0}^{m-1} (x+k)}-\dfrac1{\prod_{k=0}^{m-1} (x+1+k)}\\ &=\dfrac1{\prod_{k=0}^{m-1} (x+k)}-\dfrac1{\prod_{k=1}^{m} (x+k)}\\ &=\dfrac1{\prod_{k=1}^{m-1} (x+k)}(\dfrac1{x}-\dfrac1{x+m})\\ &=\dfrac1{\prod_{k=1}^{m-1} (x+k)}(\dfrac{x+m-x}{x(x+m)})\\ &=\dfrac{m}{\prod_{k=0}^{m} (x+k)}\\ &=\dfrac{m}{p_{m+1}(x)}\\ \end{array} $

$\begin{array}\\ q_m(n) &=\sum_{x=1}^{n} p_{m}(x)\\ &=\dfrac1{m+1}\sum_{x=1}^{n} (p_{m+1}(x)-p_{m+1}(x-1))\\ &=\dfrac{p_{m+1}(n)}{m+1}\\ r_m(n) &=\sum_{x=n}^{\infty}\dfrac1{p_{m}(x)}\\ &=\dfrac1{m-1}\sum_{x=n}^{\infty}\left(\dfrac1{p_{m-1}(x)}-\dfrac1{p_{m-1}(x+1)}\right)\\ &=\dfrac1{(m-1)p_{m-1}(n)}\\ \end{array} $

Answer 3

Starting the summations at $k=1$ and using Pochhammer symbols $$\prod_{n=0}^i(k+n)=k (k+1)_i$$ $$\sum_{k=1}^N\frac 1{k (k+1)_i}=\frac{1}{i \,\Gamma (i+1)}-\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}$$ Now, using Stirling approximation for large values of $N$

$$\log\Bigg[\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}\Bigg]=-(i \log (N)+\log (i))-\frac{i (i+1)}{2 N}+O\left(\frac{1}{N^2}\right)$$ $$\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}\sim\frac{N^{-i}}{i}\,e^{-\frac{i (i+1)}{2 N}}$$

Edit

Going from $n$ to $2n$ as asked in comments is much more difficult since the result invoke the Gaussian hypergeometric functions. $$\prod_{n=0}^i(k+2n)=2^i k \left(\frac{k}{2}+1\right)_i$$ and the summation up to $N$ gives $$S_i=\frac{\sqrt{\pi } 2^{-i-1} \, _2F_1\left(\frac{1}{2},1;i+\frac{3}{2};1\right)}{\Gamma \left(i+\frac{3}{2}\right)}+\frac{2^{-i-1} \, _2F_1(1,1;i+2;1)}{\Gamma (i+2)}-$$ $$\frac{2^{-i-1} \Gamma \left(\frac{N}{2}+\frac{1}{2}\right) \, _2F_1\left(1,\frac{N+1}{2};\frac{1}{2} (2 i+N+3);1\right)}{\Gamma \left(i+\frac{N}{2}+\frac{3}{2}\right)}-$$ $$\frac{2^{-i-1} \Gamma \left(\frac{N}{2}+1\right) \, _2F_1\left(1,\frac{N+2}{2};i+\frac{N}{2}+2;1\right)}{\Gamma \left(i+\frac{N}{2}+2\right)}$$ Once expanded, they write in a simple manner in terms of the gamma function. $$S_1=\frac{1}{4} \left(3-\frac{(N+4) \Gamma \left(\frac{N}{2}+1\right)}{2 \Gamma \left(\frac{N}{2}+3\right)}-\frac{(N+3) \Gamma \left(\frac{N+1}{2}\right)}{2 \Gamma \left(\frac{N}{2}+\frac{5}{2}\right)}\right)$$ $$S_2=\frac{1}{8} \left(\frac{11}{12}-\frac{(N+6) \Gamma \left(\frac{N}{2}+1\right)}{4 \Gamma \left(\frac{N}{2}+4\right)}-\frac{(N+5) \Gamma \left(\frac{N+1}{2}\right)}{4 \Gamma \left(\frac{N}{2}+\frac{7}{2}\right)}\right)$$ $$S_3=\frac{1}{16} \left(\frac{7}{30}-\frac{(N+8) \Gamma \left(\frac{N}{2}+1\right)}{6 \Gamma \left(\frac{N}{2}+5\right)}-\frac{(N+7) \Gamma \left(\frac{N+1}{2}\right)}{6 \Gamma \left(\frac{N}{2}+\frac{9}{2}\right)}\right)$$ $$S_4=\frac{1}{32} \left(\frac{163}{3360}-\frac{(N+10) \Gamma \left(\frac{N}{2}+1\right)}{8 \Gamma \left(\frac{N}{2}+6\right)}-\frac{(N+9) \Gamma \left(\frac{N+1}{2}\right)}{8 \Gamma \left(\frac{N}{2}+\frac{11}{2}\right)}\right)$$ where you can see the simple patterns.

Expanding the gamme functions leads to the expressions you wrote.

When $N\to \infty$, the constant term is $$T_i=2^{-i-1} \left(\frac{\sqrt{\pi } \, _2F_1\left(\frac{1}{2},1;i+\frac{3}{2};1\right)}{\Gamma \left(i+\frac{3}{2}\right)}+\frac{\, _2F_1(1,1;i+2;1)}{\Gamma (i+2)}\right)$$ that is to say $$T_i=\frac 1{2^{i+1}\,i ^2} \left(\frac{i\sqrt{\pi } }{\Gamma \left(i+\frac{1}{2}\right)}+\frac{1}{\Gamma (i)}\right)=\frac 1{2^{i+1}}\frac{1}{i \,\Gamma (i+1)}\left(1+\frac{\sqrt{\pi }\, \Gamma (i+1)}{\Gamma \left(i+\frac{1}{2}\right)}\right)$$ and then $$S_i\sim T_i-\frac {N^{-i}} i e^{-\frac{i (2 i+1)}{2 N} }$$

Answer 4

$\sum_{k=1}^{N} \frac{1}{k(k+1)\cdot\cdot\cdot(k+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)\cdot\cdot\cdot(N+i)}$ is easy to prove by induction.

It is true for $N=1$ since $\sum_{k=1}^{1} \frac{1}{k(k+1)\cdot\cdot\cdot(k+i)}=\frac{1}{(1+i)!}$ equals $\frac{1}{i\cdot i!}-\frac{1}{i\cdot 2\cdot 3\cdot\cdot\cdot(1+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot(1+i)!}= \frac{1}{(1+i)!}$

Assume the statement applies for a certain $N$. Evaluating for $N+1$: $\sum_{k=1}^{N+1}\frac{1}{k(k+1)\cdot\cdot\cdot(k+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot(N+1)\cdot\cdot\cdot(N+i)}+\frac{1}{(N+1)\cdot\cdot\cdot (N+1+i)}=$

$=\frac{1}{i\cdot i!}-\frac{N+1+i-i}{i\cdot (N+1)\cdot\cdot\cdot(N+i)}=\frac{1}{i\cdot i!}-\frac{1}{i\cdot (N+2)\cdot\cdot\cdot(N+1+i)}$ shows that it also applies for $N+1$ which ends the proof.