Submitted by marshmallow on Fri, 03/31/2017 - 13:39
Forums:
I have been stuck on part iv) of this question for a while now.
I have tried doing proof by contradiction by writing e as the fraction c/d where c and d are co-prime, but have not progressed from there.
I rearranged the equation in part iii) with c/d as the subject
I replaced An with 1/(n+p) where p is positive, since we know that 0
Sorry, It didn't fully send- This is what I meant
I have been stuck on part iv) of this question for a while now.
I have tried doing proof by contradiction by writing e as the fraction c/d where c and d are co-prime, but have not progressed from there.
I rearranged the equation in part iii) with c/d as the subject
I replaced An with 1/(n+p) where p is positive, since we know that An is less than 1/n.
I kept [n!e] as it was as I didn't know whether I could factorize out anything from it.
Thank you, appreciate any guidance or help.
I think you're slightly
I think you're slightly overcomplicating this. You assumed that $e=c/d$, so $d!e$ is then an integer. But then $[d!e] = d!e$ so what does part (iii) tell you about $a_d$? Why is this a contradiction (hint: part (ii)).
Ad = 0
Ad = e(d!)-[d!e]
this would mean Ad = e(d!)-e(d!) = 0 when e = c/d, which is a contradiction since Ad must be more than 0!
Thank you very much! :)
Perfect. :-)
Perfect. :-)
I don't really have much of a
I don't really have much of a clue on how to start this, I understand the final argument for the irrationality of e but I'm not entirely keen on the infinite series parts.
My approaches were:
Factorising out $\frac{1}{n+1}$ to show the series is self inclusive $b_n=\frac{1}{n+1}\left(\frac{1}{n+1}+ \frac{1}{(n+1)^2} +\frac{1}{(n+1)^3} +\ ...\right)$
Then $b_n=\frac{1}{n+1}b_n$ but that only shows that $1 = \frac{1}{n+1}$ is that because I'm dividing by something that, at the very least, will approach 0 as $n \to \infty$?
I also tried combining everything into a single fraction $$b_n = \frac{1 + x + x^2 +\ ...\ + x^{\infty - 1}}{x^\infty}$$
But I don't like the presence of $\infty$ in the fraction
Looking slightly further ahead leads me to believe that $b_n = \frac{1}{n}$ but I can't see a way of manipulating what I have to that, although looking at the fraction supports this is in the case that $n =1$
Am I along the right lines or do I need a different approach altogether?
Your very first step is not
Your very first step is not correct, as I'm sure you can see. If you factorise out $\frac{1}{1+n}$ you must see that you get $b_n = \frac{1}{n+1} \left(1 + \frac{1}{n+1} + \cdots \right)$, that is, the first number in the bracket is $1$, not $1/(n+1)$.
Instead, you're massively overcomplicating this. Let $r = \frac{1}{n+1}$ then you're trying to find an expression for $b_n = r+ r^2 + r^3 + \cdots $.
**cough** geometric series...
Ah yes - actually with the $b
Ah yes - actually with the $b_n = \frac{1}{n+1}(1 + b_n)$ I can rearrange to get my $b_n = \frac{1}{n}$
But I can see that using $S_\infty=\frac{a}{1-r}$ will rearrange also provided you set $a = 1$ then subtract one since that term doesn't appear or set $a = \frac{1}{n+1}$
For part (ii) is it sufficient for me to compare the terms in the sequences $a_n$ and $b_n$ and say that after the first term all of $a_n$ terms are less than the corresponding $b_n$ terms so $a_n < b_n$ then $a_n < \frac{1}{n}$?
Obviously $a_n > 0$ since you're adding positive terms
Part (iii) from multiplying the infinite series for $e^x$ where $x = 1$ by $n!$ I obtain $$n! + n! + \frac{n!}{2!}+\frac{n!}{3!} + \cdots + \frac{n!}{r!} + \cdots$$
Since n is constant and the $e^x$ series is infinite there will come a time where $ n \lt r$ - everywhere before that will provide integer values for the fractions hence the minus integer part in the equation- so letting $r! = n!(r-n)!$ at that point then the series becomes $$\frac{n!}{n!(n+1)}+\frac{n!}{n!(n+1)(n+2)}+ \frac{n!}{n!(n+1)(n+2)(n+3)} + \cdots$$
Which cancelling the $n!$ terms gives us our $a_n$ sequence as required - but I feel there might be a slightly more elegant way of putting things
Yep, your way of getting $b_n
Yep, your way of getting $b_n$ is fine. Yes, (ii) is also correct.
Your approach to (iii) is also perfect, but for the sake of clarity, I would write it like this:
$$\begin{align*}e &= 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} + \cdots \\ n!e &= n! + n! + \frac{n!}{2} + \cdots + \frac{n!}{n!} + \frac{n!}{(n+1)!} + \frac{n!}{(n+2)!} + \cdots \\ & = n! + n! + \frac{n!}{2} + \cdots + 1 + \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \cdots \end{align*}$$
which I'm sure you can see is almost the same as yours but has the benefit of not mentioning the redundant $r$ term. I expressed the series for $e$ solely in terms of $n$ and ellipses, there's no need for an extra variable $r$ that you then later set to something specific. That said, if having $r$ there helps you with the thought process, then by all means. But by large your work is currently correct and well done.
Also, I'm not sure what you
Also, I'm not sure what you mean by $r! = n!(r-n)!$, you've just defined $r$ in terms of itself, which seems sketchy to me - but your end result seems fine, so you know what you were doing, but probably just poorly expressed yourself with this whole $r$ business (which is why incidentally, I find that not introducing that redundant $r$ variable better).
Sorry, it's the triple whammy
Sorry, it's the triple whammy of confusing, wrong and redundant. My formula book has $\frac{x!}{r!}$ as the continuation of the $e^x$ series which I tried to incorporate into my answer. Partway through I realised that it didn't really work using it especially in the form $r!=n!(r−n)!$ as it implied that $n!e = [n!e] + e - 2$ as the cancelling through the fractions would have $\frac{1}{2!} + \frac{1}{3!} +\frac{1}{4!} \cdots$ making $a_n = e - 2$. Basically my definition for $r!$ was totally wrong. So I didn't use that part for the answer but forgot to remove it from my explanation
Ah no worries, as long as you
Ah no worries, as long as you understand!