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seen Jan 13 '13 at 1:25

Jul
2
awarded  Curious
Jun
11
awarded  Popular Question
Jan
12
comment Convergence of a serie depending on an exponent
Yes, it is...but thank you for the help.
Jan
11
comment Convergence of a serie depending on an exponent
I don't know, the problem is that I have to hand this exercise in in half an hour...1/n^(2k)?
Jan
11
comment Convergence of a serie depending on an exponent
Yes he wants to see a proof, but I thought the fact that I have proved the p-test for another exercice,would make it possible, that I could use this.Do you have a suggestion for an other proof?Perhaps the integral test?
Jan
11
comment Convergence of a serie depending on an exponent
I know that the majorant is in this intervall 0<k<1 divergent, is that explanation enough?
Jan
11
asked Convergence of a serie depending on an exponent
Jan
11
accepted Convergence of $\sum \limits_{k=1}^\infty (\frac{k^3}{-1-k^4})$
Jan
11
comment Convergence of $\sum \limits_{k=1}^\infty (\frac{k^3}{-1-k^4})$
Thank you that was really helpful.My fault was that I have tried to do it by a majorant which diverges...
Jan
11
revised Convergence of $\sum \limits_{k=1}^\infty (\frac{k^3}{-1-k^4})$
added 1 characters in body
Jan
11
asked Convergence of $\sum \limits_{k=1}^\infty (\frac{k^3}{-1-k^4})$
Jan
10
comment Convergence of series 1
Thanks now I see it in 3) so to do the comparison test of a fraction with 2 polynoms, I always have to neglect the other terms from lower degree in the denominator and after this I only have regard the different serie terms with the p-test?
Jan
10
accepted Convergence of series 1
Jan
10
comment Convergence of series 1
then if it works for 3), it also works for 1) and i can say that it's convergent because "1)" = $1/l^2$ and that converges?
Jan
10
comment Convergence of series 1
perhaps.. I thougth I should write the 3 series because my questions are quite connected for the 3 series.
Jan
10
comment Convergence of series 1
So I can neglect the other terms,in the same way as if I would regard the limit of a sequence?
Jan
10
asked Convergence of series 1
Dec
22
comment Do there exist sequences with these properties and these limit points?
I think now I have understood (3), thank you.
Dec
21
accepted Proofs for the sequence $a_n$ with $a_n = \cos\left(\frac{n\pi}{4}\right)$
Dec
21
comment Show that $\sum_{k=1}^\infty \frac{1}{k!}$ converges without the Ratio Test
No, it's ok.I thought perhaps one could prove it over $\sum_{k=1}^\infty \frac{1}{k!}=e=\lim\limits_{n\rightarrow\infty} (1- \frac {1}{n})^n$ but it's nearly the same.(Anyway I believe I would have to prove the convergence of the sum)