Recent content by Lisa91

How to prove that n^{\alpha} > \ln(n) for \alpha>0 ?

How to prove that for x>0 \ln(1+x) > \frac{x}{2+x} is true?

Thank you so much! It's so beautiful!

Could anyone tell me please why the limit of this guy is infinity? \lim_{n\to\infty} \frac{n!-1}{n^{3} \ln(n!)}

\ln \{\ln (13j + k) >1 \ln (13j + k)>e 13j + k >e^{e} so if we take j=1 and k=0,1,... it's true and we can also take k=9 and j=0,1... Is it ok?

|z|= |0^{2}+(-1)^{2}| = |1| so according to the rule we don't know whether it is convergent or not...

Do think wolfram alpha is wrong in this case?

Thank you! I don't have any doubts about the fact that the limit of this guy is zero \frac{\ln (13\ j + k)}{13\ j + k} . Do you think the explanation 'we may consider it as a constant term' is good for the exam? I think I feel what you mean. Indeed, it increases very slowly but still I am...

We can write the following ones as: \sum_{n=1}^{\infty}(-1)^na_{2n} -a_{1}+a_{3}-a_{5}+... \sum_{n=1}^{\infty}(-1)^na_{2n-1} -a_{2}+a_{4}-a_{6}+... So we get: -a_{1}-a_{2}+a_{3}+a_{4}-a_{5}-a_{6}+... \sum_{n=1}^{\infty}(-1)^na_{n} the series differs a liitle bit.

\lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \begin{cases} 1&x \in \mathbb{Q} \\0& x \not\in \mathbb{Q}\end{cases} No, this '2k' has to be in the place I wrote.

I tried to do it this way but I don't know how to prove that it decreases and that the limit is zero. I tried estimating it RHS and LHS using \frac{t}{t+1}< \ln(t+1) < t but in one case I've got -1... (-1)^{j}\ \frac{\ln (13\ j + 1)}{(13\ j + 1)\ \ln \{\ln (13\ j + 1)\}} .

When x = e^{2} +2 we get 2x+2< x \ln(x) . So, this is not what I want. Ok, I thought I could solve it on my own but it seems to be much more complicated. I have a series \sum (-1)^n \frac{2\ln(n)}{\sqrt{n+1}} . I thought I could use the Leibniz test but even wolfram alpha shows that the...

\lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \begin{cases} 1&x \in \mathbb{Q} \\0& x \not\in \mathbb{Q}\end{cases} . How to prove it?

I have one series \sum_{n=13}^{\infty}(-1)^{\left\lfloor\frac{n}{13}\right\rfloor} \frac{ \ln(n) }{n \ln(\ln(n)) } . How to investigate its convergence? I wanted to group the terms of this series but I don't know whether it's a good idea as we have 13 terms with minus and then 13 with plus and...

I am not so sure whether I got the idea. I divide the series into two parts. It's clear when I write its first terms. But then we take a+b - the imaginary and real part. On what basis can we add these two parts? I guess the main idea is to show that the series \sum_{n=1}^{\infty}(-1)^na_{n}...