# Thread: Series Convergence Conditionally or absolutely

1. ## Series Convergence Conditionally or absolutely

Does $\displaystyle \sum_{n=1}^\infty (-1)^n\frac{5^{2n}}{n!}$ convergence conditionally or absolutely?

How do you take the limit of a factorial?

Thanks

2. Originally Posted by polymerase
Does $\displaystyle \sum_{n=1}^\infty (-1)^n\frac{5^{2n}}{n!}$ convergence conditionally or absolutely?

How do you take the limit of a factorial?

Thanks
$\displaystyle \lim_{n \to {\infty}}\bigg|\frac{25\cdot{5^{2n}}}{(n+1)\cdot{n !}}\cdot\frac{n!}{5^{2n}}\bigg|\Rightarrow\lim_{n \to {\infty}}\bigg|\frac{25}{n+1}\bigg|=0$ which is less than one so it converges absolutely since $\displaystyle |a_{n}|$ converges

3. Originally Posted by polymerase
How do you take the limit of a factorial?
In general you cannot take the limit of a factorial.

Almost all the time, one should have some kind of an intuition for such a problem. The tests should only reinforce your suspicion.

But you have various attitudes to look at any problem. I am writing this article to help you attack the problem in various ways. Hope it helps

1) I shalt follow the rules
A sign of a learning student.Here,for instance, you can apply the tests. Good number of tricky problems can have you stumped though.

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2) I shalt investigate
A sign of a curious individual. He need not even be studying this theory. Takes long time to analyse things. Just not efficient for exams. Yet this approach is the basic, researcher's attitude. Example:

So in factorial problems like this, look at it this way: The numerator is constantly increasing 25 times, every step. But the denominator is increasing by the step number. Which means eventually it will beat this constant 25. But wait! This only tells me that the sequence will have a diminishing tail. And all have been duped by harmonic series who went on forever with his diminishing tail. We need something more to establish convergence. Well look at the pace at which the factorial grows.
When n = 26 $\displaystyle \frac1{n} < \frac1{25}$
When n = 27 $\displaystyle \frac1{n(n-1)} < \frac1{25^2}$
When n = 28 $\displaystyle \frac1{n(n-1)(n-2)} < \frac1{25^3}$
So get the clue?

Even more incredible is the fact that this intuition could have worked for any number 'a' instead of 25. Since any finite number would have been eventually reached.

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3) I shalt be aware.
This is generally the approach many teachers and the so called 'cream' have. You should simply possess more knowledge or more importantly 'identify the problem' to apply that knowledge. This is harder as it mostly comes with experience.
Example: $\displaystyle \sum_{n=1}^\infty (-1)^n\frac{5^{2n}}{n!}$
The above expression looks familiar. I have seen these factorials hanging below somewhere. Also the numerators look like powers of some numbers. Hey that is even more familiar.....
Doesnt it look like the power series for $\displaystyle e^{x}$?
Hey wait, if I get that -1 inside the power and write $\displaystyle 5^{2n}$ as $\displaystyle 25^n$, then it looks like this:

$\displaystyle \sum_{n=1}^\infty \frac{(-25)^{n}}{n!}$

Well.....Isnt this $\displaystyle e^{-25} - 1$ ?
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I am still following steps (1) and (2), but I hope one day I will reach (3) with experience

4. Originally Posted by Isomorphism
In general you cannot take the limit of a factorial.

Almost all the time, one should have some kind of an intuition for such a problem. The tests should only reinforce your suspicion.

But you have various attitudes to look at any problem. I am writing this article to help you attack the problem in various ways. Hope it helps

1) I shalt follow the rules
A sign of a learning student.Here,for instance, you can apply the tests. Good number of tricky problems can have you stumped though.

------------------------------------------------------------------------------------------------------------------

2) I shalt investigate
A sign of a curious individual. He need not even be studying this theory. Takes long time to analyse things. Just not efficient for exams. Yet this approach is the basic, researcher's attitude. Example:

So in factorial problems like this, look at it this way: The numerator is constantly increasing 25 times, every step. But the denominator is increasing by the step number. Which means eventually it will beat this constant 25. But wait! This only tells me that the sequence will have a diminishing tail. And all have been duped by harmonic series who went on forever with his diminishing tail. We need something more to establish convergence. Well look at the pace at which the factorial grows.
When n = 26 $\displaystyle \frac1{n} < \frac1{25}$
When n = 27 $\displaystyle \frac1{n(n-1)} < \frac1{25^2}$
When n = 28 $\displaystyle \frac1{n(n-1)(n-2)} < \frac1{25^3}$
So get the clue?

Even more incredible is the fact that this intuition could have worked for any number 'a' instead of 25. Since any finite number would have been eventually reached.

------------------------------------------------------------------------------------------------------------------

3) I shalt be aware.
This is generally the approach many teachers and the so called 'cream' have. You should simply possess more knowledge or more importantly 'identify the problem' to apply that knowledge. This is harder as it mostly comes with experience.
Example: $\displaystyle \sum_{n=1}^\infty (-1)^n\frac{5^{2n}}{n!}$
The above expression looks familiar. I have seen these factorials hanging below somewhere. Also the numerators look like powers of some numbers. Hey that is even more familiar.....
Doesnt it look like the power series for $\displaystyle e^{x}$?
Hey wait, if I get that -1 inside the power and write $\displaystyle 5^{2n}$ as $\displaystyle 25^n$, then it looks like this:

$\displaystyle \sum_{n=1}^\infty \frac{(-25)^{n}}{n!}$

Well.....Isnt this $\displaystyle e^{-25} - 1$ ?
------------------------------------------------------------------------------------------------------------------
I am still following steps (1) and (2), but I hope one day I will reach (3) with experience
Is my method invalid?

5. Originally Posted by polymerase
Does $\displaystyle \sum_{n=1}^\infty (-1)^n\frac{5^{2n}}{n!}$ convergence conditionally or absolutely?

How do you take the limit of a factorial?

Thanks
As post #2 by Mathstud28 shows, the factorials cancel and no limit of them is required.

Nevertheless, in general you could take the limit of a factorial by considering the limit of the gamma function: Gamma function - Wikipedia, the free encyclopedia

6. Originally Posted by mr fantastic
As post #2 by Mathstud28 shows, the factorials cancel and no limit of them is required.

Nevertheless, in general you could take the limit of a factorial by considering the limit of the gamma function: Gamma function - Wikipedia, the free encyclopedia
yes...for example say you wanted to compute $\displaystyle \lim_{n \to {\infty}}\frac{n^n}{n!}$...you could substitute it with the gamma function to get $\displaystyle \lim_{n \to {\infty}}\frac{n^n}{\int_0^{\infty}x^{n}e^{-x}}$ then using L'hopitals you would get $\displaystyle \lim_{n \to {\infty}}\frac{n^n(ln(n)+1)}{n^n\cdot{e^{-n}}}$ then through cancellation you can see the answer is ∞

7. Using Stirling's approximation for $\displaystyle n \to +\infty$ might be quicker but it requires memorizing it: $\displaystyle n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ (it's not quite different from using the Gamma function : it derives from it)

8. Originally Posted by flyingsquirrel
Using Stirling's approximation might be quicker but it requires memorizing it: $\displaystyle n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$ (it's not quite different from using the Gamma function : it derives from it)
The only thing would be that only applies when you are evaluating a limit approaching ∞ if it was approaching 0 I dont think that would apply?

9. Have you ever wanted to approximate n! when n approaches 0 ?
It's exactly the same as your integral, you take it from 0 to $\displaystyle \infty$

10. Originally Posted by Moo
Have you ever wanted to approximate n! when n approaches 0 ?
It's exactly the same as your integral, you take it from 0 to $\displaystyle \infty$
yes...if you were doing limits...?

11. Originally Posted by polymerase
Does $\displaystyle \sum_{n=1}^\infty (-1)^n\frac{5^{2n}}{n!}$ convergence conditionally or absolutely?

How do you take the limit of a factorial?

Thanks
It converges absolutely since $\displaystyle \sum_{n=1}^\infty \frac{5^{2n}}{n!}=e^{25}-1$

In fact: $\displaystyle \sum_{n=1}^\infty (-1)^n\frac{5^{2n}}{n!}=e^{-25}-1$

12. Originally Posted by Mathstud28
yes...if you were doing limits...?
Do you really want to know the limit of n! when n approaches 0 ? oO

13. Originally Posted by Mathstud28
The only thing would be that only applies when you are evaluating a limit approaching ∞ if it was approaching 0 I dont think that would apply?
I've corrected my post
Originally Posted by Mathstud28
yes...if you were doing limits...?
What does "$\displaystyle n$ approaches 0" means when $\displaystyle n$ is an integer ? $\displaystyle n$ simply takes the value 0...

14. Originally Posted by Isomorphism
In general you cannot take the [I]limit of a factorial.
I disagree with that statement.
In fact here are two examples that have direct application to series.
$\displaystyle \lim _{n \to \infty } \frac{1}{{\sqrt[n]{{n!}}}} = 0\,\& \,\lim _{n \to \infty } \frac{n}{{\sqrt[n]{{n!}}}} = \frac{1}{e}$

The Cauchy test (root test) preferred by many is directly applicable to this problem.
$\displaystyle \left( {\sqrt[n]{{\left| {\frac{{\left( { - 1} \right)^n 5^{2n} }}{{n!}}} \right|}}} \right) = \left( {\frac{{25}}{{\sqrt[n]{{n!}}}}} \right) \to 0$

Because the series is absolutely convergent it is convergent.

15. Originally Posted by flyingsquirrel
I've corrected my post

What does "$\displaystyle n$ approaches 0" means when $\displaystyle n$ is an integer ? $\displaystyle n$ simply takes the value 0...
Haha...I meant like the limit of whatever over n! as n goes to 0..not simply n!....a limit with n! somewhere in it

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