Recent content by losiu99

Yes, that's what I was talking about. Can you see how to use it to prove your formula by induction?

Are you familiar with half-angle formula for cosine?

On the second thought, it shouldn't be there. Equidistribution doesn't guarantee that much regularity:wink: The best we can say knowing that 1/3 of sin^2n is not less than 1/2 is that the whole sum is not less than 1/2\sum_{n=1}^{N/3}\frac{1}{\sqrt{n}} Edit: the actual numbers are pulled out of...

This reasoning applies only to decreasing functions. Try something which is "large" for integers, but vanishes outside of small intervals around them.

On the other hand, http://www.wolframalpha.com/input/?i=sum_%28n%3D1%29^2000+sin^2%28n%29%2Fsqrt%28n%29

Could you please elaborate on how did you do that? Mathematica says the sum of first 10K elements is almost 100. As for proving divergence: I'd try exploiting the fact that the sequence \left{\frac{n}{2\pi}\right} is equidistributed modulo 1. It means that for each subinterval, number of...

Hint: what is the minimal value of e^{\cos(x-y)} on the whole R?

l'Hospital works if you take \frac{\frac{1}{x}}{e^{\frac{1}{x^2}}}

I'd substitute t=1/x.

\sum_{n=1}^{2N}a_n=0 for every N, and so, \lim_{N\rightarrow\infty}\sum_{n=1}^{2N}a_n=\lim_{N\rightarrow\infty}0=0

Since each member of this sequence is zero, the limit is zero as well.

\sum_{n=1}^{N}a_n=(-1)^N so the series doesn't converge. On the other hand \sum_{n=1}^{2N}a_n=0\qquad\sum_{n=1}^{2N+1}a_n=-1 so both limits exist. Sorry I left it w/o explanation, I though it was this kind of problems where the answer is kinda tricky, but easy to verify.

a_n=(-1)^n

Let's use Cauchy definition of continuity: since f(u)>0, then (1/2)f(u)>0 as well. There exist a ball B around u such that, for all v\in B we have |f(v)-f(u)|<(1/2)f(u). But this implies in particular f(u)-f(v)<(1/2)f(u) and so f(v)>(1/2)f(u).

Yes. It's enough to take just [0,1] as a subset of rationals, as it is closed and bounded, yet not compact.