If $a$ is complex and $n$ is a positive integer, prove that the series converges when $a$ is a positive integer.

Question

If $a\in\mathbb{C}$ and $n\in\mathbb{Z}^+$ define $$\begin{pmatrix}a\\0\end{pmatrix}=1\quad\mbox{and}\quad\begin{pmatrix}a\\n\end{pmatrix}=\frac{a(a-1)\ldots(a-(n-1))}{n!}.$$ It can be shown by induction that the $n$th partial sum of $$\sum\limits_{n=1}^\infty (-1)^n\begin{pmatrix}a\\n\end{pmatrix}\quad\mbox{is}\quad(-1)^n\begin{pmatrix}a-1\\n\end{pmatrix}.$$ Use this fact to prove that the series converges if $a\in\mathbb{Z}^+$.

Essentially, the summation of an alternating series of $a$ choose $n$ has an $n$th partial sum of (-1)^$n$ ($a$-1) choose $n$. I'm given that it works for $a$ as a complex number, but I'm unsure how to prove it when $a$ is a positive integer...

Also, could somebody provide a proof of the partial sum formula? I want to understand how to arrive at it but I keep getting twisted in my induction. Should I attempt to prove by a different method?

Answer 1

As soon as $n > a-1$ the partial sum is zero, so the sequence of partial sums is eventually constant, hence convergent.

Or just observe that the sum itself is a finite sum. Once $n > a$ the terms all vanish.

Answer 2

I'm going to prove the sum.

Suppose that $\sum_{k=0}^n (-1)^k \binom{a}{k} =(-1)^n\binom{a-1}{n} $.

For $n=0$ this is $\binom{a}{0} =\binom{a-1}{0} $ which is true since both are $1$.

For $n=1$ this is $\binom{a}{0}-\binom{a}{1} =-\binom{a-1}{1} $ which is true since both are $-a+1$.

Suppose it is true for $n$. Then

$\begin{array}\\ \sum_{k=0}^{n+1} (-1)^k \binom{a}{k} &=\sum_{k=0}^{n} (-1)^k \binom{a}{k}+(-1)^{n+1} \binom{a}{n+1}\\ &=(-1)^n\binom{a-1}{n}+(-1)^{n+1} \binom{a}{n+1}\\ &=(-1)^n(\binom{a-1}{n}-\binom{a}{n+1})\\ &=(-1)^n\left(\dfrac{\prod_{k=0}^{n-1}(a-1-k)}{n!}-\dfrac{\prod_{k=0}^{n}(a-k)}{(n+1)!}\right)\\ &=(-1)^n\left(\dfrac{\prod_{k=0}^{n-1}(a-(k+1))}{n!}-\dfrac{\prod_{k=0}^{n}(a-k)}{(n+1)!}\right)\\ &=(-1)^n\left(\dfrac{\prod_{k=1}^{n}(a-k)}{n!}-\dfrac{\prod_{k=0}^{n}(a-k)}{(n+1)!}\right)\\ &=(-1)^n\dfrac{\prod_{k=1}^{n}(a-k)}{(n+1)!}\left(n+1-a\right)\\ &=(-1)^{n+1}\dfrac{\prod_{k=1}^{n}(a-k)}{(n+1)!}(a-n-1)\\ &=(-1)^{n+1}\dfrac{\prod_{k=0}^{n-1}(a-(k+1))}{(n+1)!}(a-n-1)\\ &=(-1)^{n+1}\dfrac{\prod_{k=0}^{n}(a-(k+1))}{(n+1)!}\\ &=(-1)^{n+1}\binom{a-1}{n+1}\\ \end{array} $

and we are done!