Chaitanya's Random Pages

October 30, 2012

An interesting sum based on factorials

Filed under: mathematics — ckrao @ 11:40 am

This post is inspired by a question in this year’s University of Melbourne Maths Olympics. We wish to find sums of the following form.

\displaystyle \frac{1}{1\times 2} + \frac{1}{2\times 3} + \frac{1}{3\times 4} + \ldots = \sum_{i=0}^{\infty} \frac{1}{(i+1)(i+2)}

\displaystyle \frac{1}{1\times 2 \times 3} + \frac{1}{2\times 3 \times 4} + \frac{1}{3\times 4 \times 5} + \ldots = \sum_{i=0}^{\infty} \frac{1}{(i+1)(i+2)(i+3)}

More generally, how do we find

\displaystyle \frac{1}{1 \times 2 \times \ldots \times k} + \frac{1}{2 \times 3 \times \ldots \times (k+1)} + \frac{1}{3 \times 4 \times \ldots \times (k+2)} + \ldots = \sum_{i=0}^{\infty} \frac{1}{(i+1)(i+2) \ldots (i + k)}\text{?}

The first sum is familiar enough to me – writing \displaystyle \frac{1}{(i+1)(i+2)} as \displaystyle \frac{1}{i+1} - \frac{1}{i+2} the sum telescopes:

\begin{aligned} \sum_{i=0}^{\infty} \frac{1}{(i+1)(i+2)} &= \lim_{N \rightarrow \infty} \sum_{i=0}^N \frac{1}{(i+1)(i+2)}\\ &= \lim_{N \rightarrow \infty} \sum_{i=0}^N \left( \frac{1}{i+1} - \frac{1}{i+2}\right )\\ &= \lim_{N \rightarrow \infty} \sum_{i=0}^N \frac{1}{i+1} - \sum_{i=1}^{N+1} \frac{1}{i+1}\\ &= \lim_{N \rightarrow \infty} \frac{1}{2} - \frac{1}{N+2} \\ &= \frac{1}{2}.\end{aligned}

To do the next sum I thought one could use the partial fraction expansion

\displaystyle \frac{1}{(i+1)(i+2)(i+3)} = \frac{1/2}{i+1} - \frac{1}{i+2} + \frac{1/2}{i+3}

and continue as in the previous case. While this is doable, it is not easily generalisable. It is simpler to write

\displaystyle \frac{1}{(i+1)(i+2)(i+3)} = \frac{1}{2}\left( \frac{1}{(i+1)(i+2)} - \frac{1}{(i+2)(i+3)} \right).

In the general case, we write

\begin{aligned} \frac{1}{(i+1)(i+2) \ldots (i + k)} &= \frac{1}{(i+2)\ldots (i+k-1)} \left( \frac{1}{(i+1)(i+k)} \right)\\ &= \frac{1}{(i+2)\ldots (i+k-1)} \cdot \frac{1}{(k-1)}\left( \frac{1}{(i+1)} - \frac{1}{(i+k)} \right)\\ &= \frac{1}{k-1} \left( \frac{1}{(i+1)(i+2)\ldots (i+k-1)} - \frac{1}{(i+2)(i+3)\ldots (i+k)} \right), \end{aligned}

leading to

\begin{aligned} \sum_{i=0}^{\infty} \frac{1}{(i+1)(i+2) \ldots (i + k)} &= \lim_{N \rightarrow \infty} \sum_{i=0}^N \frac{1}{(i+1)(i+2) \ldots (i + k)}\\ &= \lim_{N \rightarrow \infty} \frac{1}{k-1} \sum_{i=0}^N \left( \frac{1}{(i+1)(i+2)\ldots (i+k-1)} - \frac{1}{(i+2)(i+3)\ldots (i+k)} \right) \\ &= \lim_{N \rightarrow \infty} \frac{1}{k-1} \left( \sum_{i=0}^N \frac{1}{(i+1)(i+2) \ldots (i+k-1)} - \sum_{i=1}^{N+1} \frac{1}{(i+1)(i+2) \ldots (i + k - 1)} \right) \\ &= \lim_{N \rightarrow \infty} \frac{1}{k-1} \left( \frac{1}{1 \times 2 \times \ldots \times (k-1) } - \frac{1}{(N+2)(N+3) \ldots (N+k) } \right) \\ &= \frac{1}{(k-1) \times (k-1)!}. \end{aligned}

 The reason for the title of the post is this sum can also be written as

\displaystyle \sum_{i=0}^{\infty} \frac{i!}{(i+k)!} = \frac{1}{(k-1) \times (k-1)!}, \quad k \geq 2.

Multiplying both sides by k! we also obtain the following interesting infinite sum involving reciprocals of binomial coefficients:

\displaystyle \sum_{i=0}^{\infty} \binom{i+k}{k} ^{-1} = \frac{k}{k-1}, \quad k \geq 2.

For example,

\displaystyle \binom{3}{3} ^{-1} + \binom{4}{3} ^ {-1} + \binom{5}{3} ^{-1} + \binom{6}{3} ^{-1} + \ldots = \frac{1}{1} + \frac{1}{4} + \frac{1}{10} + \frac{1}{20} + \ldots = \frac{3}{2}.

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