Another 'Does this series converge?'

[Mattofix]

Homework Statement

The sum from n=1 to infinity of n!*2^n*n^(-n)

Homework Equations

The Attempt at a Solution

not a clue.

 

[masnevets]

try using the ratio test

 

[torresmido]

you should first make the series simple:

The series you have is the same as: the sum from n=1 to infiniti of
n!*(2/n)^n "because n^(-n) is the same as 1/n^(n)).

Now you can use the ratio test:
take the limite as n approaches infiniti of [(n+1)!*(2/n)^(n+1)] / [n!*(2/n)^n ]
this limit becomes after simplifying:
limit as n approaches infiniti of (2n+2)/n
you can fact the n out and cancel it out with the n in the denominator to get:
limit as n approaches infiniti of 2+(2/n) which equals 2
so because 2 is grater than one, this series converges by the ratio test

 

[masnevets]

No, when you simplify that ratio, you get 2(n/(n+1))^n, whose limit is 2/e. This is less than 1, so the series CONVERGES.

If you get 2 as the limit, the ratio test says that it DIVERGES.

 

[torresmido]

(2n+2)/n
=n*(2+2n^-1)/n
so now you cancel out the n in the denominator with the one above it
you get:
2+(2/n)

when you take the limit as n approaches infiniti you get 2

 

[Mattofix]

when you simplify that ratio, you get 2(n/(n+1))^n, whose limit is 2/e.

how do you get this?

i know that (1 + c/n)^n -> e^c as n-> infin

 

[Mattofix]

but how do you get 2/e, well c= -1, but how do you get that?

 

[torresmido]

Alright,

the series you have is n!*2^n*n^-n
this can be simplifyed to become:

An= n!*(2/n)^n

so An+1= (n+1)!*(2/n)^(n+1) "you just replace n by n+1

The ratio test says lim as n goes to infinity of An+1/An
which is the same as:

lim as n aprroaches infiniti of [(n+1)!*(2/n)^(n+1)] / [n!*(2/n)^n]

(n+1)!/n! equals n+1 and (2/n)^(n+1)/(2/n)^(n) equals 2/n

so the limit becomes:

lim as n approaches infiniti of (n+1)*(2/n)

which is the same as:

lim as n approaches infiniti of (2n+2)/n
you can factor out the

The limit becomes:

n*(2+2*n^-1)/n

cancel out the n in the denominator and the one above it:

now you have the limit as n aprroaches infiniti of 2+(2/n) or 2+2*n^-1

which equals 2

 

[Dick]

Alright,

the series you have is n!*2^n*n^-n
this can be simplifyed to become:

An= n!*(2/n)^n

so An+1= (n+1)!*(2/n)^(n+1) "you just replace n by n+1

The ratio test says lim as n goes to infinity of An+1/An
which is the same as:

lim as n aprroaches infiniti of [(n+1)!*(2/n)^(n+1)] / [n!*(2/n)^n]

(n+1)!/n! equals n+1 and (2/n)^(n+1)/(2/n)^(n) equals 2/n

so the limit becomes:

lim as n approaches infiniti of (n+1)*(2/n)

which is the same as:

lim as n approaches infiniti of (2n+2)/n
you can factor out the

The limit becomes:

n*(2+2*n^-1)/n

cancel out the n in the denominator and the one above it:

now you have the limit as n aprroaches infiniti of 2+(2/n) or 2+2*n^-1

which equals 2

Wrong again. Not, "An+1= (n+1)!*(2/n)^(n+1)". An+1=(n+1)!*(2/(n+1))^(n+1). Mattofix, can you correct the rest? As masnevets has already pointed out, the ratio test limit is 2/e.

 

[Mattofix]

yeah - i know that torresmido is wrong.

i have handed it in already though.

but i got to 2(n/(n+1))^n and then didnt know how to get 2/e as the limit.

thanks for your help guys

 

[Dick]

yeah - i know that torresmido is wrong.

i have handed it in already though.

but i got to 2(n/(n+1))^n and then didnt know how to get 2/e as the limit.

thanks for your help guys

Write it as 2/((n+1)/n)^n=2/(1+1/n)^n. Look familiar now?

 

[torresmido]

Sorry guys ... I see my mistake now.
the limimit is 2/e so the series diverges by the ratio test

 

[Dick]

Sorry guys ... I see my mistake now.
the limimit is 2/e so the series diverges by the ratio test

You mean converges, right? 2/e<1.

 

[Mattofix]

oh - yeah, now it makes sense