@Chris'ssis Apply Stolz Cesaro Lemma, then factor out $(n+1)$ in the denominator and we have a Riemann sum, right? Next exercise for me $$\lim_{n\to\infty} n^k \Big( 1 - \prod_{j=n}^{\infty} \frac{ j^{k+1} - \alpha}{ j^{k+1} + \alpha} \Big)$$ :P
@rehband you can also note something else $$1-e^{\sum}\sim -\sum$$, all being based on that elementary limit, $\displaystyle \lim_{x\to0} \frac{e^x-1}{x}=1$
@TedShifrin You are indirectly asking though. That means you are a nosy curious, intelligent old man. I am using my analytic mind to apply to Online Detection.
We do grow up faster in here. People of equivalent age elsewhere that come to visit always seem childish to us, but maybe that's just because we are snobbish
Oh, the STEP exam serves as an admissions test separate to A-Levels/IB (which are necessary for almost all universities) to get into math related courses at Cambridge, Warwick, Imperial (in some cases) and UCL (in some cases).
I think I could do straight-to-doctorate programs in here with my current grades, might as well go for that.. But still a semester away from finishing this degree
@rehband Well, it's simple: take the maximum term multiplied by $n$ for getting the bound on the right side and then consider AM-GM for the left side combined with Stolz theorem. Done.
@PedroTamaroff $$\lim_{n\to\infty}\frac{n}{e^{a_n}}=\lim_{n\to\infty}\frac{1}{e^{a_(n+1)}-e^{a_(n)}}$$ and then you replace a_{(n+1)} by the expression in the recurrence.
@PedroTamaroff Yeah, this discussion reminds me of some nice sequence I solved in the past where I found a very nice solution, a kind of generalization. Let me find it.
@Hippalectryon I solved some open problems form that book. Well, I might say anything about that book, but I wouldn't like to be misunderstood: it's really the best book I have ever read on this topic. :-)
@Hippalectryon btw, yesterday I created this one. Did you see it? $$\int_0^1\sqrt{x}\tan^{\large 1/2^2}(2\arctan(x))\tan^{\large 1/2^3}(2^2\arctan(x))\tan^{\large 1/2^4}(2^3\arctan(x))\cdots \ dx = \log(2)$$
@rehband the measure of the area where the integrand is far from $\cos(0)\cos^{-1}(0)$ is very small for big $n$, so we can just treat it as $\cos(0)\cos^{-1}(0)$
@AlexanderGruber Well, some come from the study of other series, some from my research. I think the answer is much more complex than what I provided. Some things simply come to mind in an unexplainable way.
@AlexanderGruber At the moment I'm a river of ideas, creativity, I might create lots of series, integrals and limits every day. I mean I don't put great efforts to do that, things comes in a natural way.
it'll have property X, and i'll think, "what other groups have property X?" or "what does property X do to this group?"
but then other times, i'll just be sitting around, and i'll think, "hey it would be really weird if there were groups that had property X. i wonder if there are some."
so one is like me solving something that comes at me, from an external source, the other is more like me just messing around and trying to solve my own weird idea.
