Question

summation (1 /(n^3 -n) )
from n=2 to infinity

Answer 1

I have been able to simplify it,

1/2(n+1) + 1/2(n-1) -1/n

but stuck after this

Answer 2

I already know the answer, don;t know how to reach there.
I don;t see a pattern when I plug in values of n

Answer 3

these is the harmonis series
\[H=\sum_{k=1}^{\infty}\frac{ 1 }{ k }\]

Answer 4

How about splitting it like this\[\frac{A}{n(n-1)}+\frac{B}{n(n+1)}\]

Answer 5

replace sum n=2 to inf with n=3 to inf for 1/2(n+1)
replace sum n=2 to inf with n=1 to inf for 1/2(n-1)

then "pull out" any terms less than n=3 to inf.
If that makes sense.....

Answer 6

\[\sum_{2}^{\infty}\frac{1}{2(n+1)}= \frac{1}{2}\sum_{3}^{\infty}\frac{1}{n}\]

Answer 7

oh you mean 1/n = 1/2n + 1/2n

Answer 8

I see we can represent the whole series as sigma(1/n) but won't we get stuck after that ?

Answer 9

\[\sum_{2}^{\infty}\frac{1}{2(n-1)}= \frac{1}{2}\sum_{1}^{\infty}\frac{1}{n}=\frac{1}{2}+\frac{1}{4}+ \frac{1}{2}\sum_{3}^{\infty}\frac{1}{n}\]

Answer 10

@mukushla sire thats what you meant ?
split 1/n in 1/2n + 1/2n ?

Answer 11

@phi but we do not know the finite sum of that series ?

Answer 12

\[-\sum_{2}^{\infty}\frac{1}{n}= \frac{1}{2}-\sum_{3}^{\infty}\frac{1}{n}\]

Answer 13

now combine the summations
\[\frac{1}{2}+\frac{1}{4}-\frac{1}{2}+\frac{1}{2}\sum_{3}^{\infty}\frac{1}{n}+\frac{1}{2}\sum_{3}^{\infty}\frac{1}{n}-\sum_{3}^{\infty}\frac{1}{n}\]

Answer 14

notice that the summations add to zero.

Answer 15

you are left with 1/4

Answer 16

oh yes, nice ! thank you @phi

Answer 17

how did that strike you ?

Answer 18

\[\huge \sum_{n=1}^{\infty} (\frac{ 1 }{ 2n(n+1) }+\frac{-1 }{2n(n-1)})\]
from @mukushla there seem to be a cancellıng of temrs

Answer 19

yes, both methods are brilliant! I understand both
thank you @mukushla and @phi (:

Answer 20

yw :)