Kenny's Identity

$\quad L$

$\displaystyle =\lim_{x\to0}\sum_{n=1}^{\infty}\sum_{r=0}^{\infty}\frac{(-1)^r}{n+r}\dbinom{n+r}{r}x^{n+r-1}$

$\displaystyle =\lim_{x\to0}\frac1x\sum_{n=1}^{\infty}x^n\sum_{r=0}^{\infty}\frac{(-1)^r}{n+r}\dbinom{n+r}{r}x^r$

$\displaystyle =\lim_{x\to0}\frac1x\sum_{n=1}^{\infty}x^n\sum_{r=0}^{\infty}\frac{(-1)^r}n\dbinom{n+r-1}{r}x^r$ [See (1)]

$\displaystyle =\lim_{x\to0}\frac1x\sum_{n=1}^{\infty}\frac{x^n}n\sum_{r=0}^{\infty}\dbinom{n+r-1}{r}(-x)^r$

$\displaystyle =\lim_{x\to0}\frac1x\sum_{n=1}^{\infty}\frac{x^n}n\frac1{(1+x)^n}$ [See (2)]

$\displaystyle =\lim_{x\to0}\frac1x\sum_{n=1}^{\infty}\frac1n\frac {x^n}{(1+x)^n}$

$\displaystyle =\lim_{x\to0}\frac1x\sum_{n=1}^{\infty}\frac1n\left(\frac x{1+x}\right)^n$

$\displaystyle =\lim_{x\to0}\frac1x\left(-\ln\left(1-\frac x{1+x}\right)\right)$ [See (3)]

$\displaystyle =\lim_{x\to0}\frac1x\left(-\ln\left(\frac1{1+x}\right)\right)$

$\displaystyle =\lim_{x\to0}\frac1x\ln(1+x)$

$\displaystyle =\lim_{x\to0}\ln(1+x)^{\frac1x}$ [See (4)]

$\displaystyle =\ln e$

$\displaystyle =1$

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(1): $\dfrac1{n+r}\dbinom{n+r}{r}$

$=\dfrac{(n+r)!}{n!r!(n+r)}$

$=\dfrac{(n+r-1)!}{n!r!}$

$=\dfrac1n\dfrac{(n+r-1)!}{(n-1)!r!}$

$=\dfrac1n\dbinom{n+r-1}{r}$

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(2): $(1-x)^{-n}\equiv\displaystyle\sum_{r=0}^{\infty}\dbinom{n+r-1}{r}x^r$ ( $-1<x<1$ ), substituting $x:-x$ .

This can be proven using induction.

For a non-induction but informal approach, consider $\dfrac1{1-x}=\dfrac{1-x^\infty}{1-x}=1+x+x^2+x^3+\cdots$ .

Then $\dfrac1{(1-x)^n}=(1+x+x^2+x^3+\cdots)^n$ .

Consider the coefficient of $x^r$ in this infini-nomial (binomial, trinomial, ...) expansion.

This is equivalent to choosing a non-negative number from the first bracket, then one from the second, until the n-th bracket, and then adding them to $r$ .

Consider dividing $r$ stars by $n-1$ bars instead, which is still equivalent.

There are in total $n+r-1$ objects in which $r$ are stars, hence the coefficient is $\dbinom{n+r-1}{r}$ .

See Combinations with Repetition for further information.

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(3): $\displaystyle\ln\left(\frac1{1-x}\right)=-\ln(1-x)=\sum_{n=1}^{\infty}\frac{x^n}n$ .

Proof:

$\quad\displaystyle\sum_{n=1}^{\infty}\frac{x^n}n$

$=\displaystyle\sum_{n=1}^{\infty}\int_0^xx^{n-1}\mathrm dx$

$=\displaystyle\int_0^x\sum_{n=1}^{\infty}x^{n-1}\mathrm dx$

$=\displaystyle\int_0^x\sum_{n=0}^{\infty}x^n\mathrm dx$

$=\displaystyle\int_0^x\frac1{1-x} dx$ [See (2)]

$=\displaystyle\left[\ln\left|\frac1{1-x}\right|\right]_0^x$

$=\displaystyle\ln\left|\frac1{1-x}\right|$

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(4): $\lim_{x\to0}(1+x)^{\frac1x}=e$

Such is the definition of $e$ .