Question

summation of n root n/(n+1)^2 form 1 to infinity

Answer 1

is that a question or an affirmation?

Answer 2

QUESTION. DOES IT DIVERGE OR CONVERGE AND WHAT TEST do i use'

Answer 3

this does not converge for sure

Answer 4

Its a question, i'm not sure where to start. sorry my keyboard is messed up

Answer 5

what part of math is it?

I think you can plut numbers from 0 to 10 (for example) and see how it converges

Answer 6

if it is
\[\sum_1^{\infty}\frac{n\sqrt{n}}{(n+1)^2}\]

Answer 7

is that the question?

Answer 8

its calculus 2 infinite series and thats it!!

Answer 9

does not converge. we do it with our eyeballs.

Answer 10

the denominator is a polynomial of degree 2. the numerator has "degree" 3/2

Answer 11

in order for this to converge the degree of the denominator has to be greater than one more than the degree of the numerator

Answer 12

so you subtract the numerator degree and denominator degree and use the p test to determine convergence?

Answer 13

yes that would work

Answer 14

the degree of the numerator is 1/n tho

Answer 15

also probably comparison test because this is evidently bigger than
\[\sum\frac{1}{n}\] which diverges

Answer 16

??
\[n\times \sqrt{n}=n^1\times n^{\frac{1}{2}}=n^{\frac{3}{2}}\]

Answer 17

wait since n^1/n-2<0 for all n...that would make p<1. wouldn't that diverge?

Answer 18

just to help prove your point...I just wanna know if that's is that a correct proof

Answer 19

i am afraid you have lost me. the denominator is a polynomial of degree 2. it is
\[n^2+2n+1\] the numerator is
\[n^{\frac{3}{2}}\] so even without the
\[\sqrt{n}\] part this would diverge. you have
\[\frac{n^{\frac{3}{2}}}{n^2+2n+1}\]

Answer 20

so with my eyeballs i say
\[2-\frac{3}{2}=\frac{1}{2}\] and
\[\frac{1}{2}<1\] so by comparison test (to say 1/n) this diverges

Answer 21

maybe you should write something fancier, but i am not sure what.

Answer 22

i see u must have multiplied both the numerator and denominator by n. I just theorized \[\sum_{?}^{?} n^1/n \div n^2 = \sum_{?}^{?} n^1/n-2\] plus any number form 1 to infinity in for n and the number will always be less than one. Thanks for your help!!

Answer 23

plug*

Answer 24

yw

Answer 25

I'm wrong?