Question

Determine if the Sum from infinity to n=1 of n^3/ 5^n

Answer 1

\[\sum_{n=1}^{\infty} \frac{ n^3 }{ 5^n}\]

Answer 2

you want the sum? or you want to know if it converges?

Answer 3

the sum if it convereges

Answer 4

actually just if it converges

Answer 5

ratio test will work nicely for this one i think

Answer 6

when using the ration test i have the lim-> infinity \[\frac{ (n+1)^3 }{ 5n^3 }\]

Answer 7

and I'm stuck

Answer 8

numerator and denominator are both polynomials of degree 3

Answer 9

No it's 5^n ?

Answer 10

leading coefficient of the numerator is 1, leading coefficient of the denominator is 5
limit is \(\frac{1}{5}\)

Answer 11

or with the ratio test yes.

Answer 12

in other words, pretty much from your eyeballs you see that
\[\lim_{n\to \infty}\frac{ (n+1)^3 }{ 5n^3 }=\frac{1}{5}\]

Answer 13

(n+1)^3 is not equal to n^3 + 1^3 ???

Answer 14

no it is not

Answer 15

but it is still a polynomial of degree 3 with leading coefficient 1

Answer 16

in fact it is \(n^3+3n^2+3n+1\) but all the terms of degree lower than 3 don't matter when you take the limit

Answer 17

oh so they would all go to 0 anywat

Answer 18

accept n^3

Answer 19

\[\lim{n\to \infty}\frac{n^3+3n^3+3n+1}{5n^3}=\frac{1}{5}\] right?

Answer 20

yes but that means it would diverge because an does not equal 0. The answer is converge

Answer 21

this from whatever pre-calc class you had where you knew that the horizontal asymptote of a rational function, where the degree of the numerator equals the degree of the denominator, is the ratio of the leading coefficients

Answer 22

so your limit is \(\frac{1}{5}\)and since \(\frac{1}{5}<1\) the sum converges

Answer 23

no you got two things confused

Answer 24

the TERMS themselves must go to zero, although that is not enough for the sum to converge
but the ratio test you have to check that the limit of the ratios is less than 1, not zero

Answer 25

\[\lim\frac{a_{n+1}}{a_n}<1\]

Answer 26

Ohhhhhhh ok thanks!

Answer 27

yw