## Loser66 one year ago Use identity $$\dfrac{1}{k(k+1) }= \dfrac{1}{k}-\dfrac{1}{k+1}$$ to calculate $$\sum_{k=1}^n \dfrac{1}{k(k+1)}$$ Please, help

1. Loser66

It's easy to see that it is a telescoping series. If I let $$a_k = 1/k$$ then $$a_{k+1}= 1/(k+1)$$. What is wrong with it?

2. Loser66

Hence the sum is = $$a_n - a_0 = (1/n) -1 = \dfrac{1-n}{n}$$ but it is wrong.

3. anonymous

is that sum supposed to be from $$k=1$$ to $$n$$?

4. Loser66

Yes

5. imqwerty

the series is like - 1-1/2 + 1/2 -1/3.......1/n -1/(n+1) =1-1/(n+1) =(n)/(n+1)

6. anonymous

consider: $$\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)=1-\frac12+\frac12-\frac13+\dots+\frac1n-\frac1{n+1}=1-\frac1{n+1}=\frac{n}{n+1}$$

7. Loser66

@imqwerty @oldrin.bataku Yes, it is, if we expand it. My question is why cannot I apply formula?

8. anonymous

yeah, you incorrectly collapsed the telescoping series. it actually collapses to: $$\sum_{k=1}^n\frac1k-\sum_{k=1}^n\frac1{k+1}=\left(1+\sum_{k=2}^n\frac1k\right)-\left(\sum_{k=2}^n\frac1n+\frac1{n+1}\right)=1-\frac1{n+1}$$

9. anonymous

I don't know what you mean by 'formula', there's no general formula for a telescoping series in that sense -- telescoping just means it can be rewritten as a series in which interior terms cancel out. exactly what gets left over and in what order does not really follow from a 'formula', it's dependent on the form of the series

10. anonymous

oops, that should be $$\sum_{k=2}^n\frac1k$$ both times there, not $$\sum\frac1n$$

11. Loser66

Oh, I got what is wrong with it. I am sorry for my silly. Yes, we have $$\sum_{k =1}^n (a_k - a_{k-1})= a_n -a_0$$

12. Loser66

but it is $$a_k -a_{k-1}$$ that is the term - its previous term while I interpreted as term - next term. I am sorry.