<s>[INST] 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?
 [/INST] For your question about how $\sum{\frac{2}{n(n+1)}}$ is split,
$$\frac{2}{n(n+1)} = \frac{2[(n+1)-n]}{n(n+1)}$$
$$\frac{2[(n+1)-n]}{n(n+1)} = \frac{2(n+1)}{n(n+1)} - \frac{2n}{n(n+1)}$$
$$\frac{2(n+1)}{n(n+1)} - \frac{2n}{n(n+1)} = \frac{2}{n} - \frac{2}{n+1}$$

So your question about you would know this in an exam situation, is that these type of question will have a similar type of denominator. Like in this question, if you observe the denominator $n(n+1)$ it can be split into $(n+1)-n$ which is equal to 1. So this makes splitting the fraction into two fractions easier such that the summation could be found out.

The rest of the steps are correct and the summation also probably correct. Hope you understood!
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