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Converge or diverge of factorials

  • Thread starter tnutty
  • Start date
327
1
1. Homework Statement

Determine whether the series converges or diverges.

[tex]\sum\frac{37}{n!}[/tex]


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

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 [tex]\sum\frac{1}{n!}[/tex] = [tex]\sum\frac{1}{n(n-1)!}[/tex]

=

[tex]\frac{1}{n}[/tex] *[tex]\frac{1}{(n-1)!}[/tex]

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

Is this right ?
 

Dick

Science Advisor
Homework Helper
26,258
618
That's not a p-series. A p-series looks like 1/n^p. Use a ratio test.
 
Last edited:
327
1
Don't know what a ratio test is. Isn't 1/n a p-series because

1/n = 1/n^1 ?
 
Your series isnt 1/n but 1/n!, can you see the difference?
 
327
1
yes but

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

if one term diverges, does not the other because Infinity * whatever != converge
 
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.
 
327
1
ok, so how do I go about this. Since I haven't learned ratio test yet
 
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:

[tex]n!>2^n[/tex] 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.
 
327
1
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
 
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 .
 

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