The question:
Find the value of $$\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots +2015}$$
If this is a duplicate, then sorry - but I haven't been able to find this question yet. To start, I noticed that this is the sum of the reciprocals of the triangle numbers.
Let $t_n = \frac{n(n+1)}{2}$ denote the $n$-th triangle number. Then the question is basically asking us to evaluate \begin{align} \sum_{n=1}^{2015} \frac {1}{t_n} & = \sum_{n=1}^{2015} \frac {2}{n(n+1)}\\ & = \sum_{n=1}^{2015}\frac{2}{n}-\frac{2}{n+1} \end{align}
Here's where my first question arises. Do you just have to know that $\frac {2}{n(n+1)} = \frac{2}{n}-\frac{2}{n+1}$? In an exam situation it would be very unlikely that someone would be able to recall that if they had not done a question like this before.
Moving on:
\begin{align} \sum_{n=1}^{2015}\frac{2}{n}-\frac{2}{n+1} & = \left(\frac{2}{1}-\frac{2}{2}\right) +\left(\frac{2}{2}-\frac{2}{3}\right) + \ldots +\left(\frac{2}{2014}-\frac{2}{2015}\right) +\left(\frac{2}{2015}-\frac{2}{2016}\right)\\ &= 2 - \frac{2}{2016} \\ & = \frac {4030}{2016} \\ & = \frac {2015}{1008} \end{align}
And I'm not sure if this is right. How does one check whether their summation is correct?