Question

Find the radius of
\[\sum_{n=0}^\infty 3^n x^{n!}\]
Please, help

Answer 1

Can you help? @xapproachesinfinity

Answer 2

hmm interesting one here :)

Answer 3

hmm i'm thinking that logs would be useful here somehow lol
don't know how yet

Answer 4

Lets use the ratio test:
\[\lim _{n \to \infty}\frac{3^{n+1}*|x|^{(n+1)!}}{3^n*|x|^{n!}}\]
\[\frac{3^{n+1}*|x|^{(n+1)!}}{3^n*|x|^{n!}}=3*|x|^{(n+1)!-n!}=3*|x|^{n*n!}\]

Answer 5

the last one should be \(3x^{n+1}\) but then??

Answer 6

hmm there is an error in your simplification

Answer 7

good catch old man about to say that :)

Answer 8

@xapproachesinfinity old man has old eyes hahaha.. although it is slowly catching up but it works well, right? young boy??

Answer 9

hmm thinking how can that help

Answer 10

we take the limit
and keep track of possible conditions of x lol

Answer 11

3|x|.|x|^n limit of x^n?

Answer 12

since we have the || that gotta either be 0 or infinity

Answer 13

can we have cases lol

Answer 14

Forget it. I just wonder how to work on it. It is not my real problem

Answer 15

\[\frac{3^{n+1}|x|^{(n+1)!}}{3^n|x|^{n!}}=3|x|^{(n+1)!-n!}=3|x|^{(n+1)*n!-n!}=3|x|^{n!((n+1)-1)}=3|x|^{n!*n}\]

Answer 16

wolfram said it converges where did you find the problem

Answer 17

eh that is actually right :)

Answer 18

but it doesn't help to get anywhere

Answer 19

let c be the finite number that is converges to
\[c=\sum 3^nx^{n!} \Longrightarrow \ln c=\sum \ln (3^nx^{n!}) \]
\[=\sum (n\ln 3+x\ln(n!))\]

Answer 20

some crazy nonsense here hehe

Answer 21

\[=\ln3\sum n+x\sum\ln n!\]
sum n diverges lol

Answer 22

@Loser66

Answer 23

you give up haha

Answer 24

So what you are saying is that this sum never converges regardless the value of x, right?

Answer 25

oh no! something is wrong there!
i don't even know if that's allowed to do. the series converge when i checked it in wolfram

Answer 26

@Zarkon

Answer 27

we need some insight :)

Answer 28

eh made a mistake
\[=\ln3\sum n+\ln x\sum n!\]

Answer 29

now both n! and n are divergent series! lol
that means this method won't give anything useful

Answer 30

Why cant we use the ratio test and then use diff cases for x?

Answer 31

So when |x| is less than 1 then the series converges to 0
If |x| equals 1 then the series converges to 3*1=3
And if |x| is greater than 1 then it diverges and goes to \(\infty\)

Answer 32

whoops n*n! goes to infinity as n goes to infinty

Answer 33

it does not converges o the limit is zero but the series not necessary to 0
for the first case you mentioned

Answer 34

but i think this series is beyond just a ratio test
there is a big subtlety with that limit

Answer 35

actually when the limit is zero the radius is infinite

Answer 36

in other words converges for any x

Answer 37

@SithsAndGiggles

Answer 38

The root test only works for series of the form \(\sum a_n x^n\), not \(\sum a_n x^{n!}\).

Answer 39

what about ratio
that was ratio test not root

Answer 40

I actually meant "ratio" :P but yes the same goes for ratio test.

Answer 41

i guess ratio won't work as well

Answer 42

how can we do this one

Answer 43

@Loser66 had a very similar problem before, but then we had \(x^{n^2}\). Setting \(k=n^2\) worked, but I don't think a similar substitution will work this time around.

Answer 44

what if we you some N! apprax'ts

Answer 45

use*

Answer 46

We might be able to make a fair guess about the interval of convergence, though. Clearly the series diverges if \(x=\pm1\), since that gives us a geometric series \(\sum 3^n\). I think it might be \(|x|<1\), so the radius would be \(1\).

Answer 47

I like the approximation suggestion. Stirling's approximation says
\[n!\approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n\]
I'm not sure if this will get us anywhere though. There might be some trick we can use involving logarithms.

Answer 48

That is my problem :) I don't know how to turn it to the form \(a_kx^k\)

Answer 49

that needs some real work

Answer 50

gotta leave :)

Answer 51

@SithsAndGiggles How about this problem? Find the radius of convergent series
\[\sum_{n=0}^\infty \dfrac{5^n}{n!}x^n\]

Answer 52

My other question on it is : Is there any difference if n =1 to infinitive instead of n =0?

Answer 53

The ratio test works quite nicely.
\[\lim_{n\to\infty}\left|\frac{\dfrac{5^{n+1}}{(n+1)!}x^{n+1}}{\dfrac{5^n}{n!}x^n}\right|=5|x|\lim_{n\to\infty}\frac{n!}{(n+1)!}\]
The starting index of the series does not matter.

Answer 54

So, the radius =\(\infty\) , right?

Answer 55

Right, since \(0<1\), the second series converges everywhere.

Answer 56

How about this: I would like to apply Hadamard test this time, please
\[\sum_{n=0}^\infty \dfrac{n^n}{n!}x^n\]

Answer 57

The net is so bad. :(

Answer 58

Let's use a strategy similar to the one you had before.
\[a_k=\begin{cases}3^n&\text{for }k=n!\\
0&\text{otherwise}\end{cases}\]
Then
\[\sqrt[k]{a_k}=\begin{cases}\large \sqrt[n!]{3^n}&\text{for }k=n!\\
0&\text{otherwise}\end{cases}=\begin{cases}3^{1/(n-1)!}&\text{for }k=n!\\
0&\text{otherwise}\end{cases}\]
Clearly, \(\limsup\limits_{k\to\infty}\sqrt[k]{a_k}=1\).