Does the series Sigma (n = 1 to infinity) ((n^9/9^n)) converge?

Answer 1

To solve this series you must apply the ratio test. Recall the ratio test states that if lim n -> ∞ | (a(subscript n+1) / a(subscript n) | < 1, then the series converges. In this case the resulting limit is lim n -> ∞ | [ (n+1)^9 / 9^(n+1) ] / [ n^9 / 9^n ] |. Simplifying this limit we get (1/9) * lim n -> ∞ | (n+1)^9 / n^9 |. The limit part of this equation is equal to one. This can be shown by applying l'Hôpital's rule nine times or recognizing that this the function inside of the limit is rational which means we only need to be concerned with the highest degree of both the numerator and denominator. So, (1/9) * lim n -> ∞ | (n+1)^9 / n^9 | = (1/9) * lim n -> ∞ | 1 | = 1/9 < 1. So we can conclude that this series converges! If I was unclear or you would like me to explain further, I am happy to do so!

Answer 2

sometimes trying out a few possibilities helps see where the sequence is going

or graph it with a graphing calculator

but if this is posted as a "calculus" problem skip ahead to the last couple paragraphs, which almost incomprehensible, may be exactly what you are looking for

meanwhile, try some n's, 1,2,3, ...11,12, 100

sum n^9/9^2 from n= integers from 1 to infinity

n= 1, 1^9/9^1 = 1/9 = .111...=about .11

n=2, 2^9/9^2 = 512/81= about 6.32

n=3, 3^9/9^3 = 3^2(3^2)(3^2)(3^2)(3)/3^2(3^2)(3^2) = 27

looks at first like it grows exponentially and diverges to infinity

but wait:

n=4 = 4^9/9^4 =about 39.95

maybe not, its growth is slowing down

and

n=5 = 5^9/9^5 = 33.08. it's slowing down, likely converges after all

n=6 =about 18.96

n=7 = about 8.44

n=8 = about 3.12

n=9, 9^9/9^9 = 1 definitely slowing down

n=10, 10^9/9^10 = about .29 seems to converge to 0

n=11, 11^9/9^11 = about .08

n=12, = about .02

...

n=100 100^9/9^100 = about 3.8 x 10^-78 = very close to zero, a decimal point then 77 zeros before 38

the sequence definitely converges to zero

the series converges to a sum less than at most 200 or so, add up 1/9, 6.32, 27, ... .02, above

use a graphing calculator, such as handheld TI83 calculator or online desmos

maximum value is about 40

then quickly converges to 0

the sequence clearly converges

or you can do a ratio test, integral test, l'hopital's rule, but

graph y=x^9/9^x, that's probably the most simple, fastest way to see it converges

sum = .11+6.32+27+39.95+33.08+18.96+8.44+3.12+1+.29+.08+.02

= about 139.31

the series converges to about 139.31

= about 139

actually 138.3666948 is more accurate

=about 138. the above calculation summing .11 through .02 involves rounding to 2 decimals, throwing the sum off a little

in general, if a>0, b>1, then lim n^a/b^n = 0

for 9^n/n^9 n from 1 to infinity, a=b>0, so the sequence converges

the integral test, although very tedious, involving integration by parts, is interesting

integral of n^9/9^n dn

= a constant = about 138.3487296, therefore, the series converges

the integral = exactly (4ln^9(3) + 18ln^8(3) + 72ln^7(3)+252ln^6(3)+756ln^5(3)+1890ln^4(3)+3780ln^3(3)_5670ln^2(3)+5670ln(3)+2835)/72ln^10(3)

BUT this may be the method you want, as you listed this problem under "Calculus" and this is Integral calculus to the hilt. If you were required to solve this problem with calculus, this is the method you want most.

and although the convergence is to about 138, it's slightly different for the decimals. but the close nearly identical answers are a good check, on the answer.