Recent content by Dodobird

Examine for which u \in \mathbb R the series \sum\limits_{n=1}^\infty \frac {(1+(-1)^n)^n}{n^2} |u|^{\sqrt{n}(\sqrt{n+1})} converges. What I found out so far: (1+(-1)^n) alternates between [0;2], that means that the whole series becomes zero for the even n. The interesting part are the odd n...

Okay well thanks for your help

Would be cool if some1 could check it. If I´m close to the solution. Thank you.

Hi Zondrina and all the other helpers/readers. I used the Cauchy-Condensationtest: For convergence: s_n = \sum\limits_{n=1}^\infty a_1+a_2+a_3+...+a_n with the estimate s_N=a_2+a_2+a_4+a_4+a_4+a_4+a_8+...+N*a_N =2a_2+4a_4+...+N*a_N =2(a_2+2a_4+...+ (\frac N2)*a_N)...

Is the integral test the same like the cauchy-criterion?

Thanks micromass and zondrina for your quick help I´m sorry for the misconception. I meant series, sorry about that. Well our teacher told us/me that the quotient and/or root-test aren´t helpful here because the first one shows convergence and the second one divergence. He then talked about...

Homework Statement Show that the following sequence \sum\limits_{n=1}^\infty \frac{1}{n^\alpha} for all real \alpha > 1 converges and for all real \alpha \leq 1 diverges. The Attempt at a Solution All I know is that the Abel-Summation is the only useful thing here, but I got no clue...

Yeah, that´s correct Sammy. Do you see any flaws? I´m pretty new to proofs in Mathematics and still struggle with it and still feel a little bit insecure when I got to prove something.

Thank you Sammy for your warm welcome Oh yeah, you are right. Both signs should point in the same direction. I mistakenly wrote it in the wrong way.Sorry about that. So it should be: \displaystyle \lim_{n \to \infty } \sqrt[n]{K}\ \Rightarrow \ 1 \leq \sqrt[n]{K} \leq 1 + ... Thx ;)

I thought about this and maybe this one here is more elegant than the other one, would be cool if someone could backcheck it. \forall n \text{ with } n > K >1: \sqrt[n]{K} <\sqrt[n]{n} and \lim_{n\to\infty}\sqrt[n]{1}= 1 \lim_{n\to\infty}\sqrt[n]{K} \leq \lim_{n\to\infty}\sqrt[n]{n} =...

Many thanks Fightfish for your quick reply so i can state the following: For the 3 real Sequences \sqrt[n]{1} , \sqrt[n]{K} , \sqrt[n]{n} \exists N \forall n \geq N is \sqrt[n]{1} \leq \sqrt[n]{K} \leq \sqrt[n]{n} and \lim_{n\to\infty}\sqrt[n]{1} = \lim_{n\to\infty}\sqrt[n]{n}...

Homework Statement Be K \geq 1. Conclude out of the statement that \lim_{n \to \infty } \sqrt[n]{n} = 1, dass \sqrt[n]{K} = 1 The Attempt at a Solution \lim_{n \to \infty } \sqrt[n]{K} \Rightarrow 1 \leq \sqrt[n]{K} \geq 1 + ... I got issues with the right inequality...