# Chaitanya's Random Pages

## November 27, 2011

### A few cute mathematical series

Filed under: mathematics — ckrao @ 3:11 am

From the geometric series $\sum_{k=0}^{\infty} z^k = \frac{1}{1-z}$ we can arrive at a couple of attractive-looking series:

$\displaystyle \sum_{n = 0}^{\infty} \frac{a^n}{(a+1)^{n+1}} = 1 \quad \quad (1)$

$\displaystyle \sum_{n=0}^{\infty} \left(1 - \frac{1}{a}\right)^n = a \quad \quad (2)$

For example, $\sum_{n=0}^{\infty} \frac{3^n}{4^{n+1}} = \sum_{n=0}^{\infty} \frac{4^n}{5^{n+1}} = 1$ and $\sum_{n=0}^{\infty} \left(\frac{7}{8}\right)^n = 8$.

Furthermore, from the sums $\sum_{k=1}^{\infty}k z^k = \frac{z}{(1-z)^2}$ and  $\sum_{k=1}^{\infty} k^2z^k = \frac{z(1+z)}{(1-z)^3}$ we obtain (after setting $z = a/(a+1)$)

$\displaystyle \sum_{n = 1}^{\infty} \frac{na^{n-1}}{(a+1)^{n+1}} = 1 \quad \quad (3)$

$\displaystyle \sum_{n = 1}^{\infty} \frac{n^2a^n}{(a+1)^n} = a(a+1)(2a+1) \quad \quad (4)$

This last sum reminds me of the identity $\sum_{n = 1}^a n^2 = a(a+1)(2a+1)/6$ for positive integers $a$, though $a$ appear in the sums in different places!

So for what values of $a$ are sums (1)-(4) valid (i.e. when do they converge)? For any positive integer $k$ the sum $\sum_{n=1}^{\infty} n^k z^n$ converges when $|z| < 1$, so this means sums (1), (3) and (4) converge when $|a/(a+1)| < 1$. In other words, $|a| < |a+1|$, or equivalently, $a$ is closer to 0 than to -1. Hence the real part of $a$ must be at least -1/2. Similarly, sum (2) converges when $|(a-1)/a| < 1$, which is equivalent to the real part of $a$ being at least 1/2.