The hard limits thread

Right. That's how I did it in fact. None of these are very tricky though (apart from Lord of the Flies' submission which is, as expected *too* tricky).

I had thought of approaching this by considering an iterated limit, but if we write:

$f(k,n) = \dfrac{1}{e^k}\left(1+\dfrac{k}{1!}+\dfrac{k^2}{2!}+\cdots+ \dfrac{k^n}{n!}\right)$

we seem to have:

$\displaystyle \lim_{k \to \infty} \lim_{n \to \infty} f(k,n) =1$

$\displaystyle \lim_{n \to \infty} \lim_{k \to \infty} f(k,n) =0$

so I'm not sure if we can gather anything from that.

This is a neat one; there is a quick, beautiful way of evaluating it - and also a terribly boring one.

$\dfrac{1}{e^n}\left(1+\dfrac{n}{1!}+\dfrac{n^2}{2!}+\cdots+ \dfrac{n^n}{n!}\right)\to\; ?$

By considering

$\displaystyle 2^n \sin \left(\frac{x}{2^n}\right)$

show that

$\displaystyle \frac{\sin x}{x} = \prod_{1}^{\infty} \cos \left(\frac{x}{2^i}\right)$

Neat! I should have added:

Hence deduce that

$\displaystyle \frac{2}{\pi} = \sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} \left(1 + \sqrt{\frac{1}{2}}\right)}\times \sqrt{\frac{1}{2} \left(1+ \sqrt{\frac{1}{2}\left(1 + \sqrt{\frac{1}{2}}\right)}\right)} \times \ldots$

A level students only

a) Show that if $\displaystyle \lim_{x \rightarrow c} f(x) = \lim_{x \rightarrow c} g(x) = 0$, then we can have any of the following results:

1. $\displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = a, a \in \mathbb{R}$

2. $\displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = \infty$

3. $\displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)} = -\infty$

b) What is the significance of this result when evaluating $\displaystyle \lim_{x \rightarrow c} \frac{f(x)}{g(x)}$ where $f(x), g(x)$ behave as described above?

$\displaystyle e^n = \sum_{k=0}^{\infty } \dfrac{n^k}{k!}$, so

$\displaystyle 1+\dfrac{n}{1!}+\dfrac{n^2}{2!}+\cdots+ \dfrac{n^n}{n!} = \sum_{k=0}^{\infty } \dfrac{n^k}{k!} - \sum_{k=n+1}^{\infty } \dfrac{n^k}{k!}$

$\displaystyle = e^n - \sum_{k=n+1}^{\infty } \dfrac{n^k}{k!}$

So $\displaystyle \lim_{n\rightarrow \infty} \dfrac{1}{e^n}\left(1+\dfrac{n}{1!}+\dfrac{n^2}{2!}+\cdots+ \dfrac{n^n}{n!}\right) = \lim_{n\rightarrow \infty} \dfrac{1}{e^n} \left( e^n - \sum_{k=n+1}^{\infty } \dfrac{n^k}{k!} \right)$

$\displaystyle = \lim_{n\rightarrow \infty} 1 - \lim_{n\rightarrow \infty} \dfrac{1}{e^n} \sum_{k=n+1}^{\infty } \dfrac{n^k}{k!}$

$\displaystyle = 1 - \lim_{n\rightarrow \infty} \dfrac{1}{e^n} \sum_{k=n+1}^{\infty } \dfrac{n^k}{k!}$.

All that's left is to evaluate that second limit now but it's clear that it, and hence the original limit, is between 0 and 1. Initial thoughts on evaluating it are to bound it above and below by expressions that tend to the same limit as $n \rightarrow \infty$ that are easier to find and then use Squeeze Theorem.

Will return to this later (I should go do my actual work ). This is most definitely the boring way.. ah well.

Rather than try to estimate the "error sum" directly, I think if you start from the Lagrange Remainder form of Taylor's theorem you'll get something more tractable. (I think you will still need to use Stirling's estimate for n! to finish it, so it may not be what the OP had in mind).

I'll do it tomorrow - I've had an awful headache all day so haven't got round to it yet. I don't know many hard series though - I'm more interested in seeing other people's examples.

Neat!

I should have added:

Hence deduce that

$\displaystyle \frac{2}{\pi} = \sqrt{\frac{1}{2}} \times \sqrt{\frac{1}{2} \left(1 + \sqrt{\frac{1}{2}}\right)}\times \sqrt{\frac{1}{2} \left(1+ \sqrt{\frac{1}{2}\left(1 + \sqrt{\frac{1}{2}}\right)}\right)} \times \ldots$

Looks like it's already been made...

http://www.thestudentroom.co.uk/showthread.php?p=61014269&highlight=hard%20sums

This is very nice but I think that it's pretty straightforward to get this from your original result - which is itself very nice.

Got any more like it?

That's fine by me, though I would have called it the "hard series" thread - "hard sums" sounds a little too much like what your 8 year old does as homework.

If anyone is interested, here is the neat solution I had in mind:

Roughly STEP level:

Find $\displaystyle \lim_{x \to 1} x^{\frac{x^\frac{1}{1-x}}{x-1}}$

Let $x = 1+\frac{1}{y}$ then $\displaystyle \ell = \lim_{y \to \infty} \left(1+\frac{1}{y}\right)^{- \left(1+\frac{1}{y}\right)^{-y} } = \lim_{y \to \infty} \left(1+\frac{1}{y}\right)^{- \dfrac{1}{ \displaystyle \lim_{y \to \infty} \left(1+\frac{1}{y} \right)^{y}} } = e^{-1/e}.$