Sum of Multichoose Reciprocals

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

r=1n( ⁣ ⁣(rm) ⁣ ⁣)1=mm1(1( ⁣ ⁣(mn) ⁣ ⁣)1)\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 r=1( ⁣ ⁣(rm) ⁣ ⁣)1=mm1\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, ( ⁣ ⁣(nm) ⁣ ⁣)\left(\!\!{n\choose m}\!\!\right) denotes the number of different ways to pick a bunch of mm smarties with nn distinct choices of colour, where order is not important but repetition of colours is allowed.

It can also be defined this way: ( ⁣ ⁣(nm) ⁣ ⁣)=(n+m1m)\left(\!\!{n\choose m}\!\!\right) = \binom{n+m-1}{m}

Note by Ariel Gershon
4 years, 9 months ago

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r=1( ⁣ ⁣(rm) ⁣ ⁣)1=mm1\sum_{r = 1}^{\infty} \left(\!\!{r\choose m}\!\!\right)^{-1} = \dfrac{m}{m-1}

( ⁣ ⁣(rm) ⁣ ⁣)=(r+m1m) \displaystyle \left(\!\!{r \choose m}\!\!\right) = \binom{r+m - 1}{m}

r=11(r+m1m)=(r1)!×m!(m+r1)! \displaystyle \sum_{r = 1}^{\infty} \dfrac{1}{\binom{r+m - 1}{m}} = \dfrac{(r - 1)!\times m!}{(m+r - 1)!}

=m!(r)(r+1).......(m+r1) = \dfrac{m!}{(r)(r+1).......(m + r - 1)}

1(r)(m+r1)=1m1(1r1m+r1) \dfrac{1}{(r)(m + r - 1)} = \dfrac{1}{m - 1} \left ( \dfrac{1}{r} - \dfrac{1}{m + r - 1} \right)

m!(r)(r+1).......(m+r1)=m!m1(1r(r+1)....(m+r2)1(r+1)....(m+r1))\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)

m!m1×1(m1)!=mm1 \dfrac{m!}{m - 1} \times \dfrac{1}{(m - 1)!} = \dfrac{m}{m - 1}

U Z - 4 years, 9 months ago

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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 ((rm)) \left( {r \choose m } \right) , though I might be wrong.

Calvin Lin Staff - 4 years, 9 months ago

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

We can always convert the formula into one that only involves binomial coefficients: k=mn1(km)=mm1(11(nm1))\sum_{k = m}^n \dfrac{1}{{k\choose m}} = \dfrac{m}{m-1} \left(1 - \dfrac{1}{\binom{n}{m-1}}\right)

Ariel Gershon - 4 years, 9 months ago

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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 - 4 years, 9 months ago

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@Jeet Dholakia Check out the wikis (in the practice section), and work on more problems!

Calvin Lin Staff - 4 years, 9 months ago

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