2
$\begingroup$

I want to find the sum to n terms of the following series. $$ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \frac{7}{4\times5\times6} + \cdots $$ I have found that the general term of the sum is \begin{equation} \sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)} \end{equation} so the nth term must be \begin{equation} \frac{2n - 1}{n(n+1)(n+2)} \end{equation} Now.. how should i evaluate the sum of the n terms? What is the method i must use in order to do so? Thank you in advance.

$\endgroup$
4
  • $\begingroup$ partial fractions $\endgroup$ Aug 13, 2015 at 11:49
  • $\begingroup$ This is my problem. I have done the partial fractions. But what happens then? Even with partial fractions we have "i" terms in the denominators. So it becomes like a harmonic series. How am i supposed to evaluate the sum of these? What am i missing? $\endgroup$
    – KeyC0de
    Aug 13, 2015 at 11:51
  • 1
    $\begingroup$ try telescoping them $\endgroup$ Aug 13, 2015 at 11:54
  • $\begingroup$ Hmm i have to look some things up. $\endgroup$
    – KeyC0de
    Aug 13, 2015 at 11:56

2 Answers 2

5
$\begingroup$

$$\displaystyle \frac{2n-1}{n(n+1)(n+2)} = \frac{2}{(n+1)(n+2)}-\frac{1}{n(n+1)(n+2)} = A-B$$

Now $$\displaystyle A = \frac{2}{(n+1)(n+2)} = 2\left[\frac{(n+2)-(n+1)}{(n+1)(n+2)}\right] = 2\left[\frac{1}{n+1}-\frac{1}{n+2}\right]$$

And $$\displaystyle B = \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{(n+2)-(n)}{n(n+1)(n+2)}\right] = \frac{1}{2}\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right]$$

$$\displaystyle = \frac{1}{2}\left\{\left[\frac{1}{n}-\frac{1}{n+1}\right]-\left[\frac{1}{n+1}-\frac{1}{n+2}\right]\right\}$$

Now $$\displaystyle \sum_{r=1}^{n}S_{n} = 2\sum_{r=1}^{n}\left[\frac{1}{r+1}-\frac{1}{r+2}\right]-\frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{r}-\frac{1}{r+1}\right]+\frac{1}{2}\sum_{r=1}^{n}\left[\frac{1}{r+1}-\frac{1}{r+2}\right]$$

Now Use Telescopic Sum

$\endgroup$
0
1
$\begingroup$

$$\begin{align} \sum\limits_{i=1}^n \frac{2i - 1}{i(i+1)(i+2)} &=\sum_{i=1}^n\frac {Ai+B}{i(i+1)}-\frac{A(i+1)+B}{(i+1)(i+2)}\\ &=\sum_{i=1}^n\frac{2i-\frac12}{i(i+1)}-\frac{2(i+1)-\frac12}{(i+1)(i+2)}\\ &=\frac34-\frac{4n+3}{2(n+1)(n+2)}\qquad\text{by telescoping}\\ &=\frac{n(3n+1)}{4(n+1)(n+2)}\qquad\blacksquare \end{align}$$

$\endgroup$
2
  • $\begingroup$ I solved the problem the long "classic" way. I saw the terms that were cancelling out etc. and got this result. A shorthand way such as your would be great, but i don't understand what you did at first. I'll have to see it more later. $\endgroup$
    – KeyC0de
    Aug 13, 2015 at 15:15
  • 1
    $\begingroup$ In general if the summand is of the form $P/[i(i+1)(i+2)...(i+m)]$ it can be expressed as the difference between two consecutive terms where the denominator is one degree less than that of the summand i.e. $Q/[(i+1)(i+2)...(i+m)]−Q/[i(i+1)(i+2)...(i+m−1)]$. As there is an $i$ in the numerator, we add that accordingly and assign an arbitrary coefficient to be determined. $\endgroup$ Aug 13, 2015 at 16:04

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.