Two to Tango

Notice that

$\displaystyle \frac 12 + \frac 14 + \frac 18 + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{n \ = \ 1}^{\infty} \left( \frac 12 \right)^n$

and

$\displaystyle \frac 16 + \frac {1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \cdots = \sum_{n \ = \ 1}^{\infty} \frac{1}{(n + 1)(n + 2)}$

both converge. The first sum is a geometric series with $\displaystyle |r| < 1$ and the second sum converges by direct comparison to $\displaystyle \sum_{n \ = \ 1}^{\infty} \frac{1}{n^2}$ .

Thus, we can apply linearity:

If $\displaystyle \sum_{n \ = \ 1}^{\infty} a_n$ and $\displaystyle \sum_{n \ = \ 1}^{\infty} b_n$ both converge, then $\displaystyle \sum_{n \ = \ 1}^{\infty} a_n \pm b_n = \sum_{n \ = \ 1}^{\infty} a_n \pm \sum_{n \ = \ 1}^{\infty} b_n$

Let $S$ be our desired sum. We have:

$\displaystyle \begin{aligned} S & = \frac 12 - \frac 16 + \frac 14 - \frac{1}{12} + \frac 18 - \frac{1}{20} + \cdots \\ & = \left(\frac 12 - \frac 16\right) + \left(\frac 14 - \frac{1}{12}\right) + \left(\frac 18 - \frac{1}{20}\right) + \cdots \\ & = \sum_{n \ = \ 1}^{\infty} \Bigg[\left(\frac 12\right)^n - \frac{1}{(n + 1)(n + 2)}\Bigg] \\ & = \sum_{n \ = \ 1}^{\infty} \left(\frac 12\right)^n - \sum_{n \ = \ 1}^{\infty} \frac{1}{(n + 1)(n + 2)} \end{aligned}$

If we can find the sum of both series, we'll have the value of $S$ .

Now for that first sum, we can apply the good ol' formula for finding sums of infinite geometric series, but that's boring! Instead, here's a nice proof without words I'll leave you with to show that $\displaystyle \sum_{n \ = \ 1}^{\infty} \left(\frac 12\right)^n = 1$ :

For the second sum, we'll use a telescoping series:

$\displaystyle \begin{aligned} \sum_{n \ = \ 1}^{\infty} \frac{1}{(n + 1)(n + 2)} & = \sum_{n \ = \ 1}^{\infty} \left(\frac{1}{n + 1} - \frac{1}{n + 2}\right) \\ & = \lim_{n \to \infty} \sum_{k \ = \ 1}^{n} \left(\frac{1}{k + 1} - \frac{1}{k + 2}\right) \\ & = \lim_{n \to \infty} \left(\frac 12 - \frac 13\right) + \left(\frac 13 - \frac 14\right) + \left(\frac 14 - \frac 15\right) + \cdots + \left(\frac 1n - \frac{1}{n + 1}\right) + \left(\frac{1}{n + 1} - \frac{1}{n + 2}\right) \\ & = \lim_{n \to \infty} \left(\frac 12 - \frac{1}{n + 2}\right) \\ & = \frac 12 \end{aligned}$

Hence, we have

$\displaystyle \begin{aligned} S & = \sum_{n \ = \ 1}^{\infty} \left(\frac 12\right)^n - \sum_{n \ = \ 1}^{\infty} \frac{1}{(n + 1)(n + 2)} \\ & = 1 - \frac 12 \\ & = \boxed{\displaystyle \frac 12} \end{aligned}$