Question

*Does the sum converge or diverge?

sum of ((1)/(n^2(n^2+1))) n=1 b=infinity

I got that it converges because of the p-series. simplify the denominator.

Answer 1

@agent0smith

Answer 2

\[\Large \sum_{1}^{\infty} \frac{ 1 }{ n^2(n^2+1) }\] = \[\Large \sum_{1}^{\infty} \frac{ 1 }{ n^4+n^2 }\]
that'll def converge as the terms def approach zero, and i think you're right about the p-series. They approach zero "quickly" enough/

Answer 3

yup thats what i did and also u can split up An and Bn and use the comparison test that says if Bn converges then An converges and Bn=(1/n^4) which converges....right?

Answer 4

\[\Large \sum_{1}^{\inf} \frac{ 1 }{ n^a }\] converges as long as a >=2. So 1/n^2 and any higher powers converge, 1/n and any lower doesn't.

Answer 5

awesome thats what i thought:D yay thank you!

Answer 6

:) helps to remember those... 1/n^2 or 1/n^4 or whatever converges. 1/n or 1/n^0.5 etc diverges

Answer 7

agreed lol

Answer 8

And you're right about the comparison test -

if you use Bn as 1/n^4, it def converges (since it'll converge faster than 1/n^2)

and thus 1/(n^4 + any positive number) will converge faster than 1/n^4