3

More specifically, I'm supposed to compute $\displaystyle\sum_{k=1}^{n} \frac 1 {k(k + 1)} $ by using the equality $\frac 1 {k(k + 1)} = \frac 1 k - \frac 1 {k + 1}$ and the problem before which just says that, $\displaystyle\sum_{j=1}^{n} a_j - a_{j - 1} = a_n - a_0$.

I can add up the sum for any $n$ but I'm not sure what they mean by "compute".

Thanks!

Alan Simonin
  • 1,056
papercuts
  • 1,873
  • 1
    I take "compute" to mean "write down a formula that works for any $n$ and doesn't have any summation signs or dot-dot-dots in it." – Gerry Myerson Feb 03 '13 at 01:04
  • 2
    Compute here means express the sum as a function of $n$ that does not involve a summation. – Brian M. Scott Feb 03 '13 at 01:04
  • They mean to express answer using only letter "n" and signs "1", "/","-" – Norbert Feb 03 '13 at 01:04
  • See also http://math.stackexchange.com/questions/309582/how-to-prove-sumn-i-1-frac1ii1-fracnn1, http://math.stackexchange.com/questions/286024/what-is-the-formula-for-1-1-cdot-21-2-cdot-31-3-cdot-4-ldots-1-nn1, http://math.stackexchange.com/questions/946712/proving-the-summation-formula-using-induction-sum-k-1n-frac1kk1 and maybe also a few other posts on this site. – Martin Sleziak Sep 26 '14 at 11:21

2 Answers2

5

Find a formula in terms of $n$ that can give you the sum for any $n$:

$\displaystyle\sum_{k=1}^{n} \dfrac{1}{k(k + 1)} = \sum_{k=1}^n \dfrac{1}{k} - \dfrac{1}{(k + 1)} = \sum_{k=1}^n\;\left(-\frac{1}{k+1} - \left(\frac{-1}{k}\right) \right) = \;\;\;?$

Knowing that $\displaystyle\sum_{k=1}^{n} a_k - a_{k - 1} = a_n - a_0$, just determine what $a_k$ and $a_{k - 1}$ represent in terms of your sum, and then express the sum in terms of the hint: in terms of $a_n$, and $a_0$ ($a_0$) meaning the $a_k$ term evaluated at k=0).

Edit for clarification:

$$a_k \iff -\dfrac{1}{k+1} \implies a_n = -\dfrac{1}{n+1},\;a_0 = -\dfrac{1}{0 + 1}$$

$$\implies a_n - a_0 = -\dfrac{1}{n+1} - \left(-\dfrac{1}{1}\right) = 1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}$$

amWhy
  • 209,954
  • Okay, so that's what I thought it might have meant.

    I've been struggling though. I don't see how the problem before helps me...

     – papercuts Feb 03 '13 at 01:18
  • +1 I think the summation exactly before "?" needs a parentheses. (-1/k+1--1/k), I mean. – Mikasa Feb 03 '13 at 03:25
  • thanks for the tip...I added them to ensure it's understood the entire "difference" is the summand! – amWhy Feb 04 '13 at 03:41
0

It means: find how much it sums to. In fact, you have already said everything you need to solve the problem. You only have to put 1 and 1 together to obtain 2.

Daniel Robert-Nicoud
  • 29,795