1. ## Explanation on this series please

The series that I'm working with is: (Series, n=1 -> infinity) n!/en^2

I decided that the ratio test would be my best bet, so I have come to the limit(n-> infinity) (n+1)/(e2n+1).
I know that this limit is infinity/infinity at the moment, which is indeterminate. Usually what I would do is place every term over the highest
n power, but I get confused with the e term in the denominator.

My book says that this limit = 0, which would make it convergent, I just get mixed up in the middle.

What is the right approach to figuring out this limit?

Thanks.

2. ## Re: Explanation on this series please

I have come to the limit(n-> infinity) (n+1)/(e2n+1).
In general, for any real a and b such that a > 1, $\displaystyle \lim_{n\to\infty}\frac{n^b}{a^n}=0$. For a discussion about this, see StackExchange. In particular, answer #3 starting from "we need only look at $\displaystyle n/c^n$" has a rigorous proof that $\displaystyle n/c^n\to0$ as $\displaystyle n\to\infty$ when c > 1.

The function f(n) = e^n describes a chain reaction while g(n) = n describes simple burning. I recommend playing with the graphs of e^n and n at large n to see just how much faster e^n grows.

3. ## Re: Explanation on this series please

I think I see what you're saying. So, because the denominator get's larger than the numerator so much faster, the limit can be said that it is equal to 0?

4. ## Re: Explanation on this series please

So, because the denominator get's larger than the numerator so much faster, the limit can be said that it is equal to 0?
Yes. Of course, saying that the denominator gets larger than the numerator so much faster is not a proof, but there is a rigorous proof of this fact. Perhaps the simplest proof is the one that uses L'Hopital's rule if you are allowed to do it.

5. ## Re: Explanation on this series please

Originally Posted by emakarov
Yes. Of course, saying that the denominator gets larger than the numerator so much faster is not a proof, but there is a rigorous proof of this fact. Perhaps the simplest proof is the one that uses L'Hopital's rule if you are allowed to do it.
You cannot use l'hospital's rule, it requires that the numerator and denominator be 0 in the limit. Also the function is not continuous as it is only defined for natural numbers n.

6. ## Re: Explanation on this series please

Originally Posted by Shakarri
You cannot use l'hospital's rule, it requires that the numerator and denominator be 0 in the limit.
Why do you think so?

Originally Posted by Shakarri
Also the function is not continuous as it is only defined for natural numbers n.
If $\displaystyle (x+1)/e^{2x+1}\to0$ over real numbers, then it trivially follows that $\displaystyle (n+1)/e^{2n+1}\to0$ over natural numbers.

7. ## Re: Explanation on this series please

Originally Posted by emakarov
Why do you think so?
Thats what I remember the lecturer saying
Consider the case
$\displaystyle \lim_{x->1} \frac{x^2}{x}$

By L'Hospital's rule
$\displaystyle \lim_{x->1} \frac{x^2}{x}=\lim_{x->1} \frac{2x}{1}$

Then 1=2

It may seem trivial that if it tends to zero for natural numbers then it tends to zero for real numbers but that does not remove the fact that a function is not differentiable at a point if it is not continuous at that point.

8. ## Re: Explanation on this series please

Originally Posted by Shakarri
Thats what I remember the lecturer saying
Consider the case
$\displaystyle \lim_{x->1} \frac{x^2}{x}$

By L'Hospital's rule
$\displaystyle \lim_{x->1} \frac{x^2}{x}=\lim_{x->1} \frac{2x}{1}$

Then 1=2
In this example, the limits of x^2 and x are non-zero numbers. L'Hopital's rule is indeed not applicable in such case. But it is applicable if the limits of the numerator and denominator are infinities, which are not numbers. See Wikipedia, MathWorld, PlanetMath and Springer Encyclopedia of Mathematics.

Originally Posted by Shakarri
It may seem trivial that if it tends to zero for natural numbers then it tends to zero for real numbers
No, the converse implication is trivial; this implication is false.

Originally Posted by Shakarri
but that does not remove the fact that a function is not differentiable at a point if it is not continuous at that point.
So we first prove convergence for differentiable functions $\displaystyle x+1$ and $\displaystyle e^{2x+1}$ and then use it to prove convergence on natural numbers.

9. ## Re: Explanation on this series please

Thanks. I now see that tending to +-infinity makes sense just by inverting a function that tends to zero.

Originally Posted by emakarov
No, the converse implication is trivial; this implication is false.
That was a typo. But its not really trivial if you have to go through a further proof you mention. You really shouldn't split up my sentence when the first part relies on the second.

I don't know how you can use it to prove convergence on the natural numbers but ok. You suggest that a proof involving L'hospital's rule would perhaps be simple so it looked like you didn't realise an extra step was required.