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Nov
10
comment Extend functions such that they are continuous
@martini: Oh again a typo... but nevertheless thanks.
Nov
10
comment Extend functions such that they are continuous
Oh typo. Fixed it now - may I ask you whether you can confirm my second result $b=4$ which I did determine the same way as $a$?
Nov
10
revised Extend functions such that they are continuous
fixed a signed error
Nov
10
asked Extend functions such that they are continuous
Nov
9
accepted Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
Nov
9
comment Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
@BrianM.Scott: Did you forgot to add a $1/2$ to the partial sum at the end? Nevertheless this looks more familiar and comprehensible and I will try working on that later on. - Thanks!
Nov
9
comment Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
@BrianM.Scott: We haven't even defined derivatives and integrals yet, currently attending "Calc. I" if you could say so. (Everything is different in Germany...) - However I still don't see how to determine the $1/4$ WolframAlpha has computed. By now I only know how to show, that my series converges.
Nov
9
comment Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
@BrianM.Scott: Looking up in my textbook the $p$-series is prooved (in a part we did not work through yet) with approximations $1/n^k\leq 1/n^2\leq 2/(n(n+1))$ for $k>1,k\in\mathbb{N}$ and the last term converges because the telescopic series converges. You claim, that my series I have to work with does not converge, however WolframAlpha claims that it converges to $1/4$ - am I missing something?
Nov
9
comment Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
We haven't discussed integrals yet in the context of limits and sequences and series so your hint does not help that much - so any easier way to show that? Furthermore I understand that I can show that the series will converge by your mentioned approximation, however I still don't know how to get the limit in the second step.
Nov
9
asked Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
Nov
5
accepted $\lim\limits_{n\to+\infty}\frac{2^{n^3}}{n!5^{n^2}-n^n}$ without special means
Nov
5
comment $\lim\limits_{n\to+\infty}\frac{2^{n^3}}{n!5^{n^2}-n^n}$ without special means
Oh I did some stupid mistakes - I have been able to get the desired result!
Nov
5
accepted Relation of (un)bounded (in)finite sets and $\min$/$\max$
Nov
5
comment $\lim\limits_{n\to+\infty}\frac{2^{n^3}}{n!5^{n^2}-n^n}$ without special means
My results are always converging to 0 like $1/n!$ or $1/(5^{n^2}-1)$. Do you know what I might have done wrong?
Nov
5
asked $\lim\limits_{n\to+\infty}\frac{2^{n^3}}{n!5^{n^2}-n^n}$ without special means
Nov
5
accepted $\lim\limits_{n\to\infty}\frac{n^{n+1}}{n!}$ with sandwich rule
Nov
5
answered Prove $\forall K > 0: \lim_{n\rightarrow\infty} \sqrt[n]{K} = 1$
Nov
5
asked $\lim\limits_{n\to\infty}\frac{n^{n+1}}{n!}$ with sandwich rule
Oct
30
awarded  Citizen Patrol
Oct
22
comment Is it possible to make a graph eulerian by adding exactly one node?
This lemma and Michaels answer helped me out!