Question

What is difference between them?
\(\sum_{n=0}^\infty \dfrac{n^n}{n!}x^n\) and \(\sum_{n=1}^\infty \dfrac{n^n}{n!}x^n\)
I have to find the radius of the first one. I know the radius of the second one is 1/e but wonder how they are different.
Appreciate any tips

Answer 1

well the first start at 0 and the second start at 1
so there is an extra term in the first

Answer 2

Does it affect the radius of convergence?

Answer 3

the second just starts at x

Answer 4

hmm i think not!
you do the ratio test for both and you end up with the same radius of convergence

Answer 5

well in convergence i think they have the same behavior

Answer 6

kid, I have a very convenient formula to get it without apply ratio or root test. It is Hadamard theorem, but it applies to n=1, not to n=0, that is why made this question

Answer 7

I find lim sup of \(a_n\)

Answer 8

hmm i see

Answer 9

well perhaps it applicable because of theorems conditions
but i think with other method the convergence should be the same, i think

Answer 10

mind showing how you do the entire convergence test with sup

Answer 11

One more question: If I use ratio test, what if I use \(\dfrac{a_n}{a_{n+1}}\)??

Answer 12

Ok

Answer 13

By Hadamard theorem, the radius of a convergence series is
\(R =\dfrac{1}{lim_{n\rightarrow \infty}sup\sqrt[n]{|a_n|}}\)

Answer 14

hmm i don't that's allowable since we are going by \[0<a_n<b_n\]
bn being a_n+1 here
ratio test was a follow of comparison test

Answer 15

i might be wrong :)

Answer 16

sorry perhaps im dumb but i thought 0^0 is indetermine...so n cannot be zero if that is true?

Answer 17

hmm interesting theorem
somehow it related to root test! can't tell further than that hehe
since i didn't do much about the sups

Answer 18

\[ \sqrt[n]{|a_n|}=\dfrac{n}{\sqrt[n]{n!}}=\dfrac{n}{n/e}=e\]

Answer 19

hence R = 1/e

Answer 20

@sdfgsdfgs That is why I want to make it clear. :)

Answer 21

@xapproachesinfinity need the proof of \(\sqrt[n]{n!}= n/e\)??

Answer 22

hmm interesting
but have some vague feeling to how you get n/e lol

Answer 23

hahahaha

Answer 24

@Loser66 if the first term at n=0 is indetermine, the whole summation will be indetermine as well...right? so there will be no convergence.

Answer 25

you read my mind lol
had a suspicious feeling lol

Answer 26

@sdfgsdfgs Let discuss later, let me give @xapproachesinfinity the proof , ok?

Answer 27

@Loser66 sure :) grabbing popcorns (n my lunch! brb)

Answer 28

gotta go eat lunch, write the proof i will review it

Answer 29

ok, enjoy the meal, I will jot it down

Answer 30

\[ln(\sqrt[n]{n!})=\dfrac{1}{n}ln (n!)=\dfrac{1}{2}\sum_{k=1}^\infty ln(k)\]
\[ln x\leq 2\in [1,2]\\lnx\leq 3\in [2,3]\\.....\\\int_1^n ln xdx\leq \sum_{k=1}^\infty ln(k)\]

Answer 31

\[\int_1^n lnx dx= xlnx -x|_1^n=nlnn-n+1\]
\[ln (n!)\geq nln(n) -n+1\]
on [1,2], \(ln x\geq ln1\)
on [2,3], \(ln x\geq ln2\)
....................................
\[\sum_{k=1}^n ln(x) =\sum_{k=1}^{n-1}ln k+ln n \leq \int_1^n lnx dx + lnx = nln n-n+1+ln n\]

Answer 32

\[nlnn-n+1\leq ln (n!)\leq nln n-n+1+ln n\\ln n-1+(1/n)\leq ln\sqrt[n]{n!}=(1/n)ln(n!)\leq ln n-1+(1/n)+(lnn/n)\]
\[(1/n)\leq ln\sqrt[n]{n!}-ln(n+1)\leq \dfrac{1+lnn}{n}\]

Answer 33

as n--> infinitive, the far left and far right go to 0, by Squeeze theorem, the middle one goes to 0 also

Answer 34

Hence \[e^{ln\sqrt[n]{n!}-ln(n+1)} = middle~ term\]

Answer 35

\[e^{1/n}\leq\dfrac{{\sqrt[n]{n!}}}{n/e}\leq e^{1+lnn/n}\]
hene, as n--> infinitive, the middle term goes to 1 by squeeze theorem, that gives \(\sqrt[n]{n!}=n/e\)

Answer 36

@sdfgsdfgs if n =0, then it is the \(\sum_{n=0}^\infty0^0\)

Answer 37

I do not know how to argue this case. Ha!!

Answer 38

@Loser66 nice proof above! but if n starts at 0, the first term kills the summation even thou the rest of terms converge. No way around it.

Answer 39

Ok, let me ask my prof on next Monday. If you are interesting still, remind me to give you the argument of my prof. :)

Answer 40

@Loser66 the summation beginning with n=0 is indetermine right? :)