Question

Find the interval of convergence of the series Sum of (x+2)^n)/n^n, from n=1 to infinity

Answer 1

you mean \(\Large\dfrac{(x+2)^n}{n^n}\)?

Answer 2

yes

Answer 3

\[\sum_{1}^{\infty} (x+2)^n/n^n\]

Answer 4

sorry i don't know how to do

Answer 5

i have to find a solution in 5 hours(

Answer 6

Well at least you know x=-2 is in the radius of convergence.

Answer 7

i guess that's not enough(

Answer 8

i am going to bet it converges for all \(x\)

Answer 9

Just looking at it, it appears to be smaller than e^x, so I think it obviously converges.

Answer 10

try the ratio test

Answer 11

Since \[e^{(x+2)}=\sum_{n=0}^{\infty}\frac{ (x+2)^n }{ n! } > \sum_{n=0}^{\infty}\frac{ (x+2)^n }{ n^n } \]

since (n^n)>(n!), then it has to converge.

Answer 12

that works

Answer 13

It's a little unconventional though. Probably nicer to do it some other way, but whatever works works.

Answer 14

without doing all the algebra i think ratio test gives \(\frac{x+2}{n+1}\) and since as \(n\to \infty\) you get \(0\) that means it converges for all \(x\)

Answer 15

@ridek i can write it out if you like, i just eyeballed the algebra
try it and see if that is what you get too

Answer 16

i found that it converges with any x, but proffessor told it's incorrect, that;s another kinda sums but i still have no another idae(

Answer 17

either there is a typo in the problem or your professor is mistaken

Answer 18

or i am mistaken
let me do the algebra and see what i get

Answer 19

hmm i think i am right
ratio test gives
\[\frac{(x+2)n^n}{(n+1)^{n+1}}\] if i am not mistaken
then
\[\lim_{n\to \infty}\frac{n^n}{(n+1)^{n+1}}=0\] so this sucker converges for all \(x\)
lets check wolfram

Answer 20

i got the same(

Answer 21

ok then lets go with that answer
maybe there is a typo in the question

Answer 22

okay then i'd just rewrite and reshow her my solution, hope to succeed

Answer 23

Use the e^(x+2) example I gave. It is almost undeniable that this power series is less than e^x, which is a known convergent series. Since the only thing that varies is the n^n and n!, we know that n^n is n number of n's while n! is multiplying all the numbers between 1 and n, so it has to be less since it's in the denominator.