Waste less time on Facebook — follow Brilliant.
×

Sum of Multichoose Reciprocals

I stumbled across an interesting formula involving the multichoose function. Let \(m,n\) be natural numbers with \(m > 1\), and suppose we want to find the sum of the reciprocals of the first \(n\) multichoose coefficients. After experimenting a little, I found this surprisingly elegant formula:

\[\sum_{r = 1}^{n} \left(\!\!{r\choose m}\!\!\right)^{-1} = \dfrac{m}{m-1} \left(1 - \left(\!\!{m\choose n}\!\!\right)^{-1} \right)\]

Furthermore, if we add them to infinity, the formula simplifies to \[\sum_{r = 1}^{\infty} \left(\!\!{r\choose m}\!\!\right)^{-1} = \dfrac{m}{m-1}\]

So my question is: can you think of a nice combinatorial proof of this formula (without using induction)?


For those of you who don't know, \(\left(\!\!{n\choose m}\!\!\right)\) denotes the number of different ways to pick a bunch of \(m\) smarties with \(n\) distinct choices of colour, where order is not important but repetition of colours is allowed.

It can also be defined this way: \[\left(\!\!{n\choose m}\!\!\right) = \binom{n+m-1}{m}\]

Note by Ariel Gershon
2 years, 4 months ago

No vote yet
1 vote

Comments

Sort by:

Top Newest

\[\sum_{r = 1}^{\infty} \left(\!\!{r\choose m}\!\!\right)^{-1} = \dfrac{m}{m-1}\]

\( \displaystyle \left(\!\!{r \choose m}\!\!\right) = \binom{r+m - 1}{m}\)

\( \displaystyle \sum_{r = 1}^{\infty} \dfrac{1}{\binom{r+m - 1}{m}} = \dfrac{(r - 1)!\times m!}{(m+r - 1)!}\)

\( = \dfrac{m!}{(r)(r+1).......(m + r - 1)}\)

\( \dfrac{1}{(r)(m + r - 1)} = \dfrac{1}{m - 1} \left ( \dfrac{1}{r} - \dfrac{1}{m + r - 1} \right)\)

\(\dfrac{m!}{(r)(r+1).......(m + r - 1)} = \dfrac{m!}{m - 1}\left( \dfrac{1}{r(r+1)....(m+r-2)} - \dfrac{1}{(r+1)....(m + r - 1)}\right)\)

\( \dfrac{m!}{m - 1} \times \dfrac{1}{(m - 1)!} = \dfrac{m}{m - 1}\) Megh Choksi · 2 years, 4 months ago

Log in to reply

@Megh Choksi Yes, that's the algebraic proof using telescoping series, which is in essence a form of induction.

Is there any other way of interpreting this formula? For example, if we could write it as a taylor series expansion?

A combinatorial proof seems less likely to me, because there is no clear interpretation of \( \left( {r \choose m } \right) \), though I might be wrong. Calvin Lin Staff · 2 years, 4 months ago

Log in to reply

@Calvin Lin Perhaps you're right that there's no good combinatorial proof. However the interpretation of \(\left(\!\!{n \choose m}\!\!\right)\) is the number of different multisets of size \(m\) (where order is not important but repetition is allowed) which contain elements chosen from \(\{1,2,3, ... n\}\).

We can always convert the formula into one that only involves binomial coefficients: \[\sum_{k = m}^n \dfrac{1}{{k\choose m}} = \dfrac{m}{m-1} \left(1 - \dfrac{1}{\binom{n}{m-1}}\right)\] Ariel Gershon · 2 years, 4 months ago

Log in to reply

@Calvin Lin Hi guys I wish to improve my mathematics skill , so can anyone help me with it , how do I proceed with this idea ? Jeet Dholakia · 2 years, 4 months ago

Log in to reply

@Jeet Dholakia Check out the wikis (in the practice section), and work on more problems! Calvin Lin Staff · 2 years, 4 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...