Limit of Sum

The given series can be re-written as

$\begin{aligned} \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + .... &= \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} +.... \\ \\ &= \sum_{n = 1}^{\infty} \dfrac{1}{n(n + 1)} \end{aligned}$

Using the decomposition $\dfrac{1}{k(k + 1)} = \dfrac{1}{k} - \dfrac{1}{k + 1}$ , we get that the $n$ th partial sum, $s_n$ , is

$\begin{aligned} s_n &= \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} +....+ \dfrac{1}{n(n + 1)} \\ \\ &= \left( \dfrac{1}{1} - \dfrac{1}{2} \right) + \left( \dfrac{1}{2} - \dfrac{1}{3} \right) + \left( \dfrac{1}{3} - \dfrac{1}{4} \right) + .... +\left( \dfrac{1}{n} - \dfrac{1}{n + 1} \right) \\ \\ &= 1 - \dfrac{1}{n + 1} \end{aligned}$

The sequence $\{s_n\}$ is convergent, so

$\begin{aligned} \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + .... &= \sum_{n = 1}^{\infty} \dfrac{1}{n(n + 1)} \\ \\ &= \lim_{n \rightarrow \infty} s_n \\ \\ &= \lim_{n \rightarrow \infty} \left( 1 - \dfrac{1}{n + 1} \right) = \boxed{1} \end{aligned}$

As you can see, the sum of the terms up to $n$ is $\frac{n}{n+1}$ . As we continue, we can see that the limit approaches 1. EDIT: Use the solution above. I am terrible at writing solutions.