# Another 'Does this series converge?'

• Mattofix
In summary, the given series converges by the ratio test with a limit of 2/e, as shown through simplifying and finding the limit using the ratio test.
Mattofix

## Homework Statement

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

## The Attempt at a Solution

not a clue.

try using the ratio test

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

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.

(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

masnevets said:
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

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

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

torresmido said:
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.

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.

Mattofix said:
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.

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

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

torresmido said:
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.

oh - yeah, now it makes sense

## 1. Does this series converge?

The answer to this question depends on the specific series in question. In general, a series converges if the terms of the series approach a finite limit as the number of terms approaches infinity. There are various tests that can be used to determine the convergence of a series, such as the comparison test, the ratio test, and the integral test.

## 2. What is the difference between absolute and conditional convergence?

Absolute convergence means that the series converges regardless of the order in which the terms are added. Conditional convergence means that the series only converges if the terms are added in a specific order. For example, the alternating harmonic series is conditionally convergent, while the regular harmonic series is absolutely convergent.

## 3. Can a series converge to different values?

No, a series can only converge to a single value. If a series converges, it means that the terms of the series approach a finite limit, which is the value to which the series converges. However, different series can converge to the same value.

## 4. What is an infinite series?

An infinite series is a sum of an infinite number of terms. It can be written in the form of Σan, where n represents the number of terms and an represents the nth term of the series. An infinite series can either converge or diverge, depending on the behavior of its terms.

## 5. How do you know if a series diverges?

A series diverges if the terms of the series do not approach a finite limit as the number of terms approaches infinity. There are various tests that can be used to determine the divergence of a series, such as the divergence test, the integral test, and the comparison test.

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