Question

I have to find out why the serie 1/(n*(n+1)) is convergent. (further explanation inside)

Answer 1

I have tried separating it like this: \[\frac{ 1 }{ n^2} + \frac{ 1 }{ n }\]

Answer 2

1) compare to \(\frac{1}{n^2}\) will do it

Answer 3

they are not equal

Answer 4

you can separate it using partial fractions, which will actually give you the sum

Answer 5

\[\frac{ 1 }{ n(n+1) } \neq \frac{ 1 }{ n^2 } + \frac{ 1 }{ n }\]

Answer 6

?

Answer 7

yes that is true, they are not equal

Answer 8

ok, now i get what i did wrong. So the argument for saying that is convergent would be ?

Answer 9

\[\frac{1}{n^2}+\frac{1}{n}=\frac{n+1}{n^2}\]

Answer 10

\[\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}\] is correct

Answer 11

Yes, I understand why I cannot separate de denominator, but now I'm trying to understand why the serie is convergent

Answer 12

reason that it converges is by comparison test
you can say \(\frac{1}{n(n+1)}<\frac{1}{n^2}\) and as long as you know that
\[\sum_{n=1}^\infty\frac{1}{n^2}\] is finite, then so is your sum

Answer 13

hmmm, I understand now

Answer 14

how would I calculate the sum ? Making an limit to the infinit ?

Answer 15

infinite sorry

Answer 16

\[\lim_{n \rightarrow \infty } \frac{ 1 }{ n(n+1) } = \frac{ 1 }{ \infty } = 0\] ?It converges at 0 ?