Find the value of $\frac{1}{1}+\frac{1}{1+2}+\frac{1}{1+2+3}+\ldots + \frac{1}{1+2+3 +\ldots+2015}$ 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?


And I'm not sure if this is right. How does one check whether their summation is correct?

Replace "2015" with "10", do it by hand, and see if your results match with your general formula.
