Recent content by nonaa

Thank you :)

Here x is of order 11, right?

I proved that G has an element of order 11, but subgroup... :confused:

But I cannot prove this, because I can't prove M is a group.

2. I have to prove that if x, y are from M, then (xy)11=e. If the group was abelian, it's easy - then (xy)11=x11.y11, but now (xy)11=xyxyxy...xy and don't know how to prove it's equal to e. 3. I'm not sure what's the term in English, but what I understand by the word "inverse" is such an...

Given a group G of order 22 and M = \{x \in G | x^{11}=e\}. Prove that M is a normal subgroup of G.I have troubles proving M is subgroup of G. If M was a subgroup, then I can show it is normal, but how to prove it's a subgroup? I know I have to show it's closed under multiplication and opposite...

Sadly, I know nothing about the digamma function :(

yes, I already found that, but how does it help us? H - harmonic series: H(n)=\sum_{k=1}^{n}\frac{1}{k}; It's not part of the problem, just thought it can be useful...

Any ideas?

Homework Statement \lim_{n\to\infty}\frac{\ln 2 - \sum_{k=1}^{n}\frac{1}{k+n}}{\ln 2 - \sum_{k=1}^{2n}\frac{(-1)^{k-1}}{k}}=? Homework Equations H(2n)-H(n)= \sum_{k=1}^{n}\frac{1}{k+n} The Attempt at a Solution I tried to use that \ln 2 = \sum^{\infty}_{k=1} \frac{(-1)^{k+1}}{k}...

Hi, I'm here for almost a day but you people are so nice and helpful :) I'm glad I've found this forum :) And how to do this? I've never solved such type of problems before. Would you give me a little hint?

For what values of M does this integral converge? \int_{-\infty}^{0}\frac{x.e^x}{(1-e^{2x})^M}dx My pathetic attempt for solution ended here: \frac{2x.e^x}{2.(-1)^M(e^{2x}-1)^M}.x^{M-1}\stackrel{x\rightarrow0}{\longrightarrow}\frac{(-1)^M}{2} M-1<1, M<2 Am I going in the right...

Ok, this is the last one, I promise :blushing: \sum_{n=0}^{\infty}(-1)^n\frac1{2n+1}=? It's a little confusing for me because there is no x. I tried with 1^n = x^n and finding the sum \sum_{n=0}^{\infty}\frac{(-x)^n}{2n+1}=? but no result...

Thank you :)

Would you tell me if my solution to the problem is correct? Again, we are searching for the sum. f(x)=\sum_{n=0}^{\infty}\frac{(3n+1)x^{3n}}{n!}=? \int\sum_{n=0}^{\infty}\frac{(3n+1)x^{3n}}{n!}dx=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{n!}=x.\sum_{n=0}^{\infty}\frac{(x^{3})^n}{n!}=xe^{x^3}...