3
$\begingroup$

I have the series $$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\cdots +\frac{1}{n\times(n+1)}$$

I know the following formulas: $$1+2+3+\cdots +n=\frac {n (n+1)}{2}\tag1$$ $$1^2+2^2+3^2+\cdots +n^2=\frac{n(n+1)(2n+1)}{6}\tag2$$ $$1^3+2^3+3^3+\cdots +n^3=\left(\frac{n(n+1)}{2}\right)^2\tag3$$ But none of $(1)(2)$ and $(3)$ worked.

Please help___.

$\endgroup$

marked as duplicate by Martin Sleziak, Adam Hughes, TastyRomeo, Jyrki Lahtonen Jan 10 '17 at 21:45

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

11
$\begingroup$

Try to observe that $$\frac{1}{n\times (n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ $\therefore$ The given series can be written as $$1-\frac12+\frac12-\frac13+\frac13+\cdots -\frac{1}{n}+\frac1n-\frac{1}{n+1}$$ Each term will cancel out other term except $1$ and $\frac{1}{n+1}$ .

$\therefore$ $$=1-\frac1{n+1}$$ $$=\frac{n}{n+1}$$ Hope it helps!!!

$\endgroup$
  • 1
    $\begingroup$ :-P I know I can't type this fast on phone, so I just wait to up vote all these LoL. $\endgroup$ – Simply Beautiful Art Jan 9 '17 at 15:58
  • 1
    $\begingroup$ @HarshKumar Thanks for the help. +1 $\endgroup$ – kovid vishesh Jan 9 '17 at 16:04
  • $\begingroup$ @kovidvishesh. Thanks to you $\endgroup$ – Harsh Kumar Jan 9 '17 at 16:20
  • 1
    $\begingroup$ @HarshKumar Nice Answer. +1 $\endgroup$ – user401699 Jan 10 '17 at 16:37
2
$\begingroup$

Hint: We can write $$\frac {1}{n(n+1)} =\frac {1}{n} -\frac {1}{n+1} $$ Can you take it from here?

$\endgroup$
1
$\begingroup$

Generally, for $a_n = a_0+(n-1)d$,$\frac{1}{a_n a_{n+1}}=\frac{1}{d}(\frac{1}{a_n}-\frac{1}{a_{n+1}})$.

Similarly, higher order fractions $\frac{1}{a_n a_{n+1} a_{n+2}} =\frac{1}{2d}(\frac{1}{a_n a_{n+1}} - \frac{1}{a_{n+1}a_{n+2}})$.

When seeing these kinds of sequence series, you can have a try to split items cancel each other.

$\endgroup$

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