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# Does this series converge? watch

1. Sum (n=1) to infinity: n! / (3^n).

I did the ratio test and got that it to equal (n+1)/3, and as n tends to infinity, we see that this series therefore diverges. But in my answers (as this was a worksheet I'm going through again for revision) I put that it diverges absolutely and therefore it converges conditionally. First, when I tried to work it out again, I thought it converged as 3^n would reach 0 quicker than n! would reach infinity right?
2. It diverges; you don't even need to use the ratio test, you can just notice that the summand doesn't tend to zero.

(Original post by claret_n_blue)
First, when I tried to work it out again, I thought it converged as 3^n would reach 0 quicker than n! would reach infinity right?
I take it you mean there? And the rate at which that goes to zero is the same as the rate at which goes to infinity... but why would go to infinity faster than ? As soon as , you multiply by 3 to get to , but you multiply by to get to , and so n! grows faster (asymptotically) than .
3. for n=1 to n=10 that sum is equal to about 90.9

for n=1 to n=20 that sum is equal to about 822292356.3

My guess is it diverges as n tends to infinity.
4. (Original post by nuodai)
It diverges; you don't even need to use the ratio test, you can just notice that the summand doesn't tend to zero.

I take it you mean there? And the rate at which that goes to zero is the same as the rate at which goes to infinity... but why would go to infinity faster than ? As soon as , you multiply by 3 to get to , but you multiply by to get to , and so n! grows faster (asymptotically) than .
Yeah, I meant 1/(3^n), oh ok, see my confusion was that I thought 1/(3^n) tended to 0 quicker than n! tended to infinity. If that's not the rule, then is there a rule saying something like exponents tend to infinity quicker than powers, or is that the other way round?
5. (Original post by Jack-)
for n=1 to n=10 that sum is equal to about 90.9

for n=1 to n=20 that sum is equal to about 822292356.3

My guess is it diverges as n tends to infinity.
Sadly that's not quite good enough. For example:

and

...which is a lot bigger... but the sum converges (to approximately ).

However the OP's sum does not converge.

(Original post by claret_n_blue)
Yeah, I meant 1/(3^n), oh ok, see my confusion was that I thought 1/(3^n) tended to 0 quicker than n! tended to infinity. If that's not the rule, then is there a rule saying something like exponents tend to infinity quicker than powers, or is that the other way round?
Have you ever studied asymptotics? (Like big-O notation). If not, you should, because it's quite useful at allowing you to spot things like this.

You can think of it a bit like this:

(where and )

That is, logarithms grow slower than powers <1, which grow slower than linear terms, which grow slower than polynomials, which grow slower than exponentials (NB you can replace the '2' with an 'e' or whatever), which grow slower than factorials, which grow slower than numbers raised to their own power, which grow slower than exponential factorials... and so on.

EDIT: The table on this Wikipedia page should help.

EDIT II: Also note that in the chain of inequalities above, I'm abusing notation. What I really mean to say is that they hold "for sufficiently large ". For example, if you put then it doesn't hold, but it does for larger n.
6. (Original post by claret_n_blue)
Yeah, I meant 1/(3^n), oh ok, see my confusion was that I thought 1/(3^n) tended to 0 quicker than n! tended to infinity. If that's not the rule, then is there a rule saying something like exponents tend to infinity quicker than powers, or is that the other way round?
a^n "beats" n^k (for any a > 0, k > 0).

But n! is neither of these forms, and n! in fact beats both of them.
7. (Original post by nuodai)

You can think of it a bit like this:

I would have thought that

By the logic you used earlier, regarding the next term:

8. (Original post by Give)
I would have thought that

By the logic you used earlier, regarding the next term:

Whoops yes, that's what I meant.

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