Loser66
  • Loser66
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
Mathematics
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SOLVED
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katieb
  • katieb
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Loser66
  • 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?
Loser66
  • Loser66
Hence the sum is = \(a_n - a_0 = (1/n) -1 = \dfrac{1-n}{n}\) but it is wrong.
anonymous
  • anonymous
is that sum supposed to be from \(k=1\) to \(n\)?

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Loser66
  • Loser66
Yes
imqwerty
  • imqwerty
the series is like - 1-1/2 + 1/2 -1/3.......1/n -1/(n+1) =1-1/(n+1) =(n)/(n+1)
anonymous
  • anonymous
consider: $$\sum_{k=1}^n\left(\frac1k-\frac1{k+1}\right)=1-\frac12+\frac12-\frac13+\dots+\frac1n-\frac1{n+1}=1-\frac1{n+1}=\frac{n}{n+1}$$
Loser66
  • Loser66
@imqwerty @oldrin.bataku Yes, it is, if we expand it. My question is why cannot I apply formula?
anonymous
  • anonymous
yeah, you incorrectly collapsed the telescoping series. it actually collapses to: $$\sum_{k=1}^n\frac1k-\sum_{k=1}^n\frac1{k+1}=\left(1+\sum_{k=2}^n\frac1k\right)-\left(\sum_{k=2}^n\frac1n+\frac1{n+1}\right)=1-\frac1{n+1}$$
anonymous
  • 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
anonymous
  • anonymous
oops, that should be \(\sum_{k=2}^n\frac1k\) both times there, not \(\sum\frac1n\)
Loser66
  • Loser66
Oh, I got what is wrong with it. I am sorry for my silly. Yes, we have \(\sum_{k =1}^n (a_k - a_{k-1})= a_n -a_0\)
Loser66
  • 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.

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