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Proving the n=3,4 AM-GM is almost immediate from the n=2 AM-GM. As I am quite sure you would not be allowed to assume AM-GM so I would just do the above. (unless the question explicitly says "hence"). Any other proof is just AM-GM in disguise.

Make sure you understand why things are true rather than just learning how to do questions.

Re: MX2 Integration Marathon Cool just saw your edit. Yep, the 1/n factor screamed L'Hospital's to me. It is not completely trivial that we can bring the limit inside the integral, but either the monotone convergence theorem or the dominated convergence theorem lets us do this just fine.

Re: MX2 Integration Marathon \lim_{t\rightarrow 1}\int_{\sqrt{t}}^t \frac{x}{x^n-1}\, dx=\lim_{t\rightarrow 1}\int_{1/2}^1 \frac{t^{2s}\log(t)}{t^{ns}-1}\, ds =\int_{1/2}^1 \lim_{t\rightarrow 1} \left(\frac{t^{2s}\log(t)}{t^{ns}-1}\right)\, ds\\ = \int_{1/2}^1 \frac{ds}{ns} =\frac{1}{n}\log(2)...

Re: MX2 Integration Marathon Oh that sounds fun, will have a look in a bit.

Re: MX2 Integration Marathon \frac{1}{2}\log(2). (logs being natural of course.)

+1 Pretty much my view.

I think that "history & philosophy" is a much better tag than "science" for HPSC but neither is really adequate.

Of course you could, seems kind of excessive to me though. With each one, the gain to your accuracy/speed would be less and less, whilst still costing you the same amount of time to do. That said, if you have this time to spare and you think you think there is more room for improvement in speed...

Re: HSC 2013 4U Marathon In a class consisting of n mathematics students, every student belongs to exactly one friendship group and each friendship group has at least 2 students. Let S(n) be the number of possible arrangements of friendship groups. Find a recursion relation for S(n) and...

Mmmm delusions of grandeur. I read a really funny story about a professor back in the day when FLT got a ton of amateur attention. He would reply to manuscripts with: "I confess that I am not qualified enough to evaluate this proof, I refer you to the expert: blah" where blah was the name of...

because of laziness and other things to do: mi goreng with 2 eggs poached inside and a red bull. ugh, so much fucking regret. will need to come up with something legit and wholesome for dinner.

No worries, glad to help :).

A crank/crackpot is basically someone who makes claims of immense academic achievements (they exist in every discipline, seemingly most of all in physics and mathematics) such as the the discovery of an elementary proof of Fermat's Last Theorem. An old form of procrastination for me was...

No, stay away from infinitely nested things in the HSC unless they are given to you! This is just an induction question. u_{n+1}=\sqrt{3u_n}>\sqrt{u_n^2}=u_n from the earlier result that u_n<3, as well as the fact that from the definition it is clear that each u_n is positive.

Re: HSC 2013 4U Marathon No-one else seems to be answering it, so: P_N=\prod_{n=2}^N \frac{n-1}{n+1} \cdot \prod_{n=2}^N \frac{(n+1/2)^2+3/4}{(n-1/2)^2+3/4}=\frac{2}{N(N+1)}\cdot \frac{(N+1/2)^2+3/4}{3}. by telescoping the two products separately. As N tends to infinity, this...

Re: HSC 2013 4U Marathon A quick (though crude) way to check your answers to things like this is chuck in n=1.

Short answer: Most guys wouldn't mind. Worry less about details. I think its a refreshing and attractive quality for a girl to know what she wants and make her intentions clear. But at the same time I don't mind doing a bit of pursuing. In the extreme though, really shy girls and loud girls...

Stop blaming your race/financial status/everything else under the sun for your current luck with women. I have seen some truly deadbeat guys pull women they really didn't deserve.

Re: HSC 2013 4U Marathon Yeah your equations are pretty similar to mine. Did mine quite carefully so am pretty sure its right. Will try the 3d one in a sec.