Question

HELP PLEASE

Answer 1

@zepdrix @518nad

Answer 2

Would #14 be \[\frac{ 1 }{ 3 }\sum_{n=1}^{\infty} \frac{ 1 }{ n }\] Is that why is diverges?

Answer 3

yeah

Answer 4

its a harmonic series

Answer 5

What is Harmonic Series?
Is it that the exponent of the term determines its outcome?

Answer 6

remember (1/n)^p, p from be greater than 1, for it to converge

Answer 7

Right. okay

Answer 8

I actually have no clue on how to do #15 tho.

Answer 9

harmonic series is this
1/1 + 1/2 + 1/3+1/4+..

Answer 10

if that diverges then so does any factor of it as thats infinity

Answer 11

Oh okay

Answer 12

For #15 Can you just multiply by (1/n^2)?
\[\sum_{n=5}^{\infty}\frac{ 6 }{ n^2-1 }=\frac{ \frac{ 6 }{ n^2 } }{ \frac{ n^2 }{ n^2 } -\frac{ 1 }{ n^2 }}= \frac{ 0 }{ 1 } = 0 ?\]

Answer 13

That tells you that the `sequence` converges to zero,
which is necessary if we have any hope of our `sum` converging.

It doesn't really tell us anything about the sum though, right?

Answer 14

Yessir

Answer 15

6/((n+1)(n-1))

Answer 16

Why would you need to split the deniminator