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