mathimatikoi.org

Replied by ex-member aziiri : \(1<a_n<2\) : we have \(1<a_1<2\). Suppose \(1<a_n<2\) then \(1<{\sqrt{2}\,}^{1}<a_{n+1}={\sqrt{2}\,}^{a_n}<{\sqrt{2}\,}^2=2\). \((a_n)_{n\geq 0} \) is increases. On \([1,2)\) define \(f(x)={\sqrt{2}\,}^x-x\) then \(f'(x)= 2^{x/2-1} \ln 2 -1 \leq f'(2)=\ln 2 -1 <0\) th...

Replied by ex-member aziiri : Take the \(\ln\) to get : \[\frac{1}{n}\sum_{k=1}^{n} \ln \frac{k}{n} \to \int_0^1 \ln(x) \ \mathrm{d}x =-1\] There are other methods such as Stolz lemma : if \(\frac{a_{n+1}}{a_n} \to l\in \mathbb{R}\) then \(\sqrt[n]{a_n} \to l\) for positive sequence, applying it gi...

Replied by ex-member aziiri : There is a problem, it should be \(h\to 0^{+}\) \[\int_{-a}^a \frac{h }{h^2 +x^2 } \ \mathrm{d}x \overset{h x =t}{=}\int_{-ah^{-1}}^{a h^{-1}} \frac{\mathrm{d}t}{t^2+1} = 2\arctan \frac{a}{h}\] Take \(h\to 0^{+}\) to get \(\pi\). Since \(f\) is continuous on a compact,...

Replied by ex-member aziiri : (a) We prove that the integral over \((0,+\infty)\) does not exist : \[\int_1^{\infty} \sin^2 \left ( \pi\left ( x+\frac{1}{x} \right ) \right ) \ \mathrm{d}x \overset{x+x^{-1}=t}{=} =\frac{1}{2} \int_2^{\infty} \sin^2 (\pi t) \left(1+\frac{t}{\sqrt{t^2-4}} \right) \ \...

Replied by ex-member aziiri : Replied by ex-member aziiri : Note that \(\sqrt{x-\sqrt{x-b}}\) is defied on \((0,+\infty)\) iff \(b=0\). In that case you can show easily that the limit of the integrand as \(x\to \infty\) is \(-\infty\). Which means that the integral diverges. Good morning.. and than...

Replied by ex-member aziiri:

Note that \(\sqrt{x-\sqrt{x-b}}\) is defied on \((0,+\infty)\) iff \(b=0\). In that case you can show easily that the limit of the integrand as \(x\to \infty\) is \(-\infty\).
Which means that the integral diverges.

Replied by ex-member aziiri : Consider \(g(z)=\frac{e^{i z}}{z^2+a^2}\), and let \(C_r\) be a the half circle \(\{re^{i t} | 0\leq t\leq \pi\}\) with \(r>a\), then : \[\int_{C_r \cup [-r,r]} g(z) \ \mathrm{d}z = 2\pi i \text{Res}(g,ia)= \frac{2\pi i e^{-a}}{2ia}= \frac{\pi e^{-a} }{a}\] And by the ...

Replied by ex-member aziiri : \[\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} x \sin \frac{1}{x} \overset{t=x^{-1}}{=} \lim_{t\to +\infty} \frac{\sin t}{t}=0.\] Therefore \(f\) is continuous at \(0\) and easily we see that it is continuous on \([0,1]\). By Heine's theorem we find that it is uniformly cont...

Replied by ex-member aziiri : First, let us take the sum up to \(m\) instead of \(\infty\). \[\sum_{n=1}^m \frac{1}{n} - \log\left(\frac{n+1}{n} \right)= \sum_{n=1}^m \frac{1}{n} -\sum_{n=1}^m \bigl(\log(n+1)-\log(n)\bigr).\] Telescoping the last sum gives us that the requested sum is : \[\lim_{m\t...

Replied by ex-member aziiri : Using MVT for the \(\ln\) function we get : \[ \ln(k+1)-\ln k\leq\frac{1}{k}\leq\ln k-\ln(k-1) .\] Now we do the sum from \(k=n+1\) to \(m n\) (telescopic sum), to get : $$\ln(mn+1)-\ln (n+1) \leq \sum_{k=n+1}^{m n} \frac{1}{k} \leq \ln m n - \ln n =\ln m.$$ It is obvi...