MHB Binomial coefficient challenge

lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
Prove the following identity:\[\sum_{n =1}^{\infty }\frac{1}{\binom{n+r}{r+1}}=\frac{r+1}{r},\: \: \: \: r,n \in \mathbb{N}.\]
 
Mathematics news on Phys.org
Hint:

Consider the difference:

\[\frac{1}{\binom{n+r-1}{r}}-\frac{1}{\binom{n+r}{r}}\]
 
Suggested solution:

We have:

\[\frac{1}{\binom{n+r-1}{r}}-\frac{1}{\binom{n+r}{r}}=\frac{r!(n-1)!}{(n+r-1)!}-\frac{r!n!}{(n+r)!} \\\\ =r!(n-1)!\left ( \frac{n+r}{(n+r)!}-\frac{n}{(n+r)!}\right ) = \frac{r!(n-1)!r}{(n+r)!}\]

Now, compare the last term with:

\[\frac{1}{\binom{n+r}{r+1}}=\frac{(r+1)!(n-1)!}{(n+r)!}\]

If we divide by $r$ and multiply by $r+1$, we have the identity:

\[\frac{1}{\binom{n+r}{r+1}}=\frac{r+1}{r}\left ( \frac{1}{\binom{n+r-1}{r}}-\frac{1}{\binom{n+r}{r}} \right )\]

Summing over all possible $n$ (the RHS is a telescoping sum):

\[\sum_{n=1}^{\infty }\frac{1}{\binom{n+r}{r+1}}=\frac{r+1}{r}\sum_{n=1}^{\infty }\left ( \frac{1}{\binom{n+r-1}{r}}-\frac{1}{\binom{n+r}{r}} \right ) \\\\ =\frac{r+1}{r}\left ( \frac{1}{\binom{r}{r}}-\frac{1}{\binom{r+1}{r}}+\frac{1}{\binom{r+1}{r}} -\frac{1}{\binom{r+2}{r}} + \frac{1}{\binom{r+2}{r}} - ... \right ) \\\\ =\frac{r+1}{r}.\]
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top