Question

would the series comparison for Sigma notation going from 2 to infinity of 1/n(n^2 - 1)^1/2 be equal to 1/n^2?

Answer 1

or how am I able to find if the equation converges or diverges?

Answer 2

it converges I think !

Answer 3

you know about Riemmann's series ?!

Answer 4

yes

Answer 5

I tried using series comparison but it didn't work cause i don't know if I'm comparing it with the right one

Answer 6

okk compare it with \[1/n^(3/2) \]

Answer 7

your serie is \[\sum_{}^{} 1/\sqrt{n(n^2-1)}\] ???!!

Answer 8

no, my serie is \[\sum_{2}^{infinity} 1/x \sqrt{x^2 -1}\]

Answer 9

yeaaah ! compare it wi 1/x^2
\[1/x(\sqrt{x^2-1})\approx( infinity)1/x^2\]
\[\lim_{x \rightarrow infinity}1/x \sqrt{x^2-1}/(1/x^2)=1\]

Answer 10

I tried it with the integral test, and after a while of playing around I managed to get:

\[ \int_2^\infty =\frac{1}{x\sqrt{x^2-1}} = \tan^{-1}(\sqrt{x^2-1}) \]
which converges for large values of x.

Answer 11

i compared it with that same one and by means of p-series it converges because p=2 which \[2>1\] and then proceeded to prove it by inequalities but I'm confused with the result \[(x^2<0)\]

Answer 12

yeaaah ; I see...but inequalities...how ?

Answer 13

based on the formulaf you do it by the integral test its not
tan−1
instead its
sec−1

Answer 14

because you have to prove that if the one you compare it with converges than the original also converges

Answer 15

yeaaah...1/x^2 you said that it converges .?! you have a problem with that ?

Answer 16

Moni I understand your point, but you can derive my formula and it should work.

Answer 17

or maybe I am really missing an important step, I won't exclude that.

Answer 18

\[ f(x) = \tan^{-1}(\sqrt{x^2-1}) \\
f'(x) = \frac{1}{x^2} \cdot \frac{1}{2\sqrt{x^2-1}} \cdot 2x = \frac{1}{x\sqrt{x^2-1}} \]

Answer 19

its ok don't worry about it guys! i managed to find it THANKS so much!!! i diet click the points together....

Answer 20

w8 , tell me how you diid it ?! please ! :)

Answer 21

by doing the inequalities \[1/x \sqrt{x^2-1}< 1/x^2\]

Answer 22

sorry but it's false ! :) it's true only if x is a negatif number !

Answer 23

then you work for each side making it \[x^2< x \sqrt{x^2-1}\] and than you square each side to eliminate the square root... and you have \[x^4< x^2(x(2-1)\] you multiply and then cancel the x^4 and the result will be \[0<-x^2\]

Answer 24

the second part should be \[x^4<x^2(x^2-1)\]

Answer 25

no !
\[x^2-1\le x^2\]
then \[\sqrt{x^2-1} \le \sqrt{x^2}=x \]
you multiply by x for both sides because x is a positive number . am I right ?!

Answer 26

if you want to do it with inequalities think about something else...

Answer 27

let me keep trying and if not i'll com back... thanks though

Answer 28

you are a real fighter ! I'm thinking about it too...:D

Answer 29

@monibea you know what does"is asymptotically equivalent to" mean ?! because with inequalities...things gets more complicated !