7
$\begingroup$

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?

$\endgroup$
  • 4
    $\begingroup$ That is correct. $\endgroup$ – Robert Z Nov 26 '17 at 9:01
  • 1
    $\begingroup$ 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. $\endgroup$ – kahen Nov 26 '17 at 9:03
  • 1
    $\begingroup$ 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. $\endgroup$ – S. Dolan Nov 26 '17 at 9:04
  • $\begingroup$ 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. $\endgroup$ – StephenG Nov 26 '17 at 9:07
  • 1
    $\begingroup$ 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. $\endgroup$ – Lord Shark the Unknown Nov 26 '17 at 10:06
1
$\begingroup$

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!

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.