# 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?

• That is correct. – Robert Z Nov 26 '17 at 9:01
• You don't have to know that $\frac2{n(n+1)} = \frac2n - \frac2{n+1}$, but you do have to know the concept of telescoping sums/series. Once you suspect that this is one of those, then it's relatively easy to work out the identity and compute the sum. – kahen Nov 26 '17 at 9:03
• This is an excellent solution. This is a reasonable exam question because it is a standard example of the method of differences that students will have met before. With regard to checking, it is really just a case of checking that every line of your argument is correct and then you can be confident about your conclusion. – S. Dolan Nov 26 '17 at 9:04
• 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. The trick of splitting a fraction up like that is a common one and useful in many situations. – StephenG Nov 26 '17 at 9:07
• You say "it would be very unlikely that someone would be able to recall that if they had not done a question like this before". Right, so the trick is to do questions like this before doing the exam. – Lord Shark the Unknown Nov 26 '17 at 10:06

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.