Converge or diverge of factorials

[tnutty]

Homework Statement

Determine whether the series converges or diverges.

\sum\frac{37}{n!}


I will just forget about the 37, and think of it as \sum\frac{1}{n!}

I can try to decompose the n!

n! = n(n-1)!
n! = n ( n-1) (n-2)!
n! = n(n-1)(n-2)(n-3)...2*1


so \sum\frac{1}{n!} = \sum\frac{1}{n(n-1)!}

=

\frac{1}{n} *\frac{1}{(n-1)!}

since 1/n is a p-series and diverges so does the series.

Is this right ?

 

[Dick]

That's not a p-series. A p-series looks like 1/n^p. Use a ratio test.

 

[tnutty]

Don't know what a ratio test is. Isn't 1/n a p-series because

1/n = 1/n^1 ?

 

[sutupidmath]

Your series isn't 1/n but 1/n!, can you see the difference?

 

[tnutty]

yes but

its 1/n * 1 / (n-1)!

if one term diverges, does not the other because Infinity * whatever != converge

 

[sutupidmath]

Well, nope!

1/n*1/n^2=1/n^3. this series obviously converges, by applying p-test

Eventhough the first one alone(the harmonic series diverges)

Moreover: 1/n*1/n both alone diverge but when multiplied together they converge , 1/n^2 according to p-test again.

 

[tnutty]

ok, so how do I go about this. Since I haven't learned ratio test yet

 

[sutupidmath]

I assume you have learned the comparison tests, right? since this is one of the first tests you learn when you are introduced to numerical series.

first prove that:

n!>2^n for some n>k, where k is a positive integer, and then use this fact and the comparison theorme to show that your series converges.

 

[tnutty]

From using a graphical interface k = 4.

not sure how I can prove it.

I could try this :(induction)

0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120

2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
2^5 = 32

 

[sutupidmath]

well, since n!>2^n, for say n>2, then it follows that

1/n!<(1/2)^n, and you probbably know that the RHS is a geometric sequence whose ratio is r=1/2<1, so it converges, now from the comparison test, we know that the series sum(1/n!) converges as well .