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10

Using fractional decomposition we get: $$\frac 1{\left(\frac{k(k+1)}{2}\right)^2} = \frac 4{(k + 1)^2} + \frac 8{(k + 1)} +\frac 4 {k^2} - \frac 8 k$$ So, your sum simplifies to one telescoping sum, and the sum $$\sum_{k=1}^\infty \frac 1 {k^2} =\frac{\pi^2} 6$$ So, all together we get: $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\dots+\... 10 The series$$\sum_{m=1}^\infty \frac{\sqrt m}{m+1}$$does not converge so can I say \lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n} does not exist? Because Zacky deleted his answer, I'll repeat the useful observation that divergence of a numerator does not mean the fraction necessarily diverges... ... 8 You've confused sequence convergence (the nth term has a finite n\to\infty limit) with series convergence (the sum of the first n terms has a finite n\to\infty limit). 5$$\frac{4^{n+1}x^{2^{n+1}}}{4^nx^{2^n}}=4x^{2^n}$$ensures convergence for all$$|x|<1$$and divergence for$$|x|\ge1.$$4 Hint:$$\sqrt{n+1}-\sqrtn=\frac1{\sqrt{n+1}^2+\sqrt{n+1}\sqrtn+\sqrtn^2}.$$4 By the theorem of Cesàro-Stolz (a discrete version of the l'Hopital rule for \frac\infty\infty), the quotient \frac{\sum_{m=1}^n a_m}{\sum_{m=1}^n b_m} has a limit if the denominator grows to infinity and the quotient \frac{a_n}{b_n} of the last terms has a limit, and then both limits have the same value. Here$$ \frac{a_n}{b_n}=\frac{\frac{\sqrt{n}}{...

4

$$\lim_{n\rightarrow\infty }\frac{\frac{1}{2}+\frac{\sqrt 2}{3}+\dots+\frac{\sqrt n}{n+1}}{\sqrt n}=\lim_{n\rightarrow\infty }\sum_{r=1}^{n}\frac{\sqrt r}{(r+1)\sqrt n}$$ $$=\lim_{n\rightarrow\infty }\sum_{r=1}^{n}\frac{\sqrt{\frac rn}}{n\frac{(r+1)}{n}}$$ $$=\lim_{n\rightarrow\infty }\frac 1n\sum_{r=1}^{n}\frac{\sqrt{\frac rn}}{\frac rn + \frac1n}$$ $$=\... 3 Remember the event$$ \{\limsup_{n\to\infty}\lvert Y_n-Y\rvert=0\} $$is$$ \{\omega\in\Omega : \limsup_{n\to\infty}\lvert Y_n(\omega)-Y(\omega)\rvert=0\}. $$But the limsup of events$$ \limsup_{n\to\infty}\{\lvert Y_n-Y\rvert=0\} $$is$$ \{\omega\in\Omega : \lvert Y_n(\omega)-Y(\omega)\rvert=0 \text{ infinitely often}\} $$which is not the same. For ... 3 Just written for your curiosity. As showed in answers, the infinit sum is not to hard. We could also have a good approximation of the partial sum$$S_p=\sum_{k=1}^p\frac{4}{k^2 (k+1)^2}$$Using a CAS,$$S_p=\frac{4 \left((p+1)^2 \left(\pi ^2-6 \psi ^{(1)}(p+1)\right)-3 p (3 p+4)\right)}{3 (p+1)^2}$$and using the asymptotics of the polygamma function$$...

3

$$\frac {1}{k(k+1)}= \frac {1}{k}-\frac {1}{k+1}$$ $$\frac {1}{k^2(k+1)^2}= \frac {1}{k^2}+\frac {1}{(k+1)^2} - \frac {2}{k(k+1)}$$ $$\sum_1^{\infty}\frac {1}{k^2(k+1)^2}= \sum_1^{\infty}\frac {1}{k^2}+\sum_1^{\infty}\frac {1}{(k+1)^2} -\sum_1^{\infty} \frac {2}{k(k+1)}=$$ $$\frac {\pi ^2}{6} +(\frac {\pi ^2}{6}-1)-2$$ $$\sum_1^{\infty} \frac {4}{k^2(k+1)... 3 For the first question, note that you have a telescoping sum, so you can write down explicitly the partial sums$$ \sum_{n=0}^N \left(\sqrt{n+1}-\sqrtn\right)=\sqrt{N+1}-\sqrt{0}=\sqrt{N+1} $$which \to\infty as N\to\infty. For the second sum, estimate$$ \sqrt{n^2+1}-\sqrt{n^2}=\frac{(n^2+1)-n^2}{(\sqrt{n^2+1})^2+\sqrt{n^2}\...

3

It follows from the ratio test that your series converges if $\lvert x\rvert<\frac16$ and diverges if $\lvert x\rvert>\frac16$. If $x=\frac16$, your series becomse $\displaystyle\sum_{n=1}^\infty\frac{n+(-3)^n}{\sqrt n}\frac1{3^n}$, which converges (it's the sum of two convergent series). And if $x=-\frac16$, your series becomse $\displaystyle\sum_{n=... 3 In fact$\left(1+\frac1n\right)^n<e$, and this can be used to prove by induction that$\left|e^n\frac{n!}{n^n}\right|>1$for all$n\geq 1$. Thus when$|x|=e$the terms don't tend to zero and it can't converge. 3 By the Cauchy-Hadamard theorem, $$\frac{1}{R}=\limsup_{n\to\infty}\sqrt[n]{|c_n|},\quad c_n=a^{n^2},\tag{1}$$ where$R$is the radius of convergence and we use the convention in this context that$\frac{1}{\infty}=0$and$\frac{1}{0}=\infty$. As you observed, $$\sqrt[n]{|c_n|}=|a|^n.$$ Thus $$\limsup_{n\to\infty}\sqrt[n]{|c_n|}=\lim_{n\to \infty}|a|^... 2$$x^n\log x^n=nx^n\log x$$and also \;nx^n=x\left(x^n\right)'\; , so for \;|x|<1\; , we get$$\sum_{n=0}^\infty x^n=\frac1{1-x}\implies \left(\sum_{n=0}^\infty x^n\right)'=\sum_{n=0}^\infty nx^{n-1}=\frac1{(1-x)^2}\implies\sum_{n=0}^\infty x^n\log x^n=\log x\left(x\sum_{n=0}^\infty \left(x^n\right)'\right)=\log x\cdot \frac x{(1-x)^2}=\log x^{... 2 If the Maclaurin series converges, but does not represent the function, then nothing can be deduced about the Taylor series, even if the radius of convergence of the Maclaurin series is infinite. Consider the standard example $$f(x)=\cases{e^{-1/x^2}, &x\neq0\\ 0, &x=0}$$ It's not hard to show that all the coefficients of the ... 2$\sum_{n=0}^N \sqrt{n+1} - \sqrt{n}$is just$\sqrt{N+1}-\sqrt 0$, because intermediate terms cancel. This tends to infinity, so the sum diverges. 2 $$P\left(\bigcup_{m\ge1}\lim\sup_{k\to > \infty}\{|X_k-X|\ge\frac{1}{m}\}\right)=0$$ Which is equivalent to$\underline{\exists\epsilon>0} $such that $$P\left(\lim\sup_{k\to \infty}\{|X_k-X|\ge\epsilon\}\right)=0$$ (Note I fixed a misplaced brace in the first statement, and changed$n$to$k$which I think is what you mean.) Those are not ... 2 The first expression concerns a$\limsup$of functions. The second expression concerns a$\limsup$of sets (which again is a set). $$\omega\in\{\limsup_{n\to\infty}|Y_n-Y|=0\}\iff\limsup_{n\to\infty}|Y_n(\omega)-Y(\omega)|=0\iff\lim_{n\to\infty}Y_n(\omega)=Y(\omega)\tag1$$ and:$$\omega\in\limsup\{|Y_n-Y|=0\}\iff\{n\in\mathbb N|Y_n(\omega)-Y(\omega)|=0\}\... 2 If f_n \to f almost everywhere define g(x)=\lim f_n(x) if the limit exist and 0 otherwise. Then we can show that g is measurable and f=g a.e.. Hence f and g are equal as elements of L^{2}. Completeness of the measure space is not required for this argument. 2 The radius of convergence is 1.$$\frac1R=\limsup_{n\to\infty}\sqrt[n]{\sum_{k=1}^n\frac1k}=\limsup_{n\to\infty}\sqrt[n]{\ln(n)+\gamma+o(1)}=\limsup_{n\to\infty}\sqrt[n]{\ln(n)},$$where \gamma is the Euler-Mascheroni constant. This is equal to 1 because$$\lim_{n\to\infty}\ln(n)^{1/n}=\lim_{n\to\infty}\exp\left(\frac{\ln(\ln n)}{n}\right)=\exp\left(\... 2 You have correctly interpreted the notation. You can use the ratio test to find the radius of convergence: $$\lim\limits_{n\to\infty} \frac{|a_n|}{|a_{n+1}|} = \lim\limits_{n\to\infty} \frac{\sum_{k=1}^n \frac{1}{k}}{\sum_{k=1}^{n+1} \frac{1}{k}} = \lim\limits_{n\to\infty} \frac{\sum_{k=1}^{n+1} \frac{1}{k} - \frac{1}{n+1}}{\sum_{k=1}^{n+1} \frac{1}{k}} = \... 2 Note that$$2\epsilon\int_{0}^{\frac{\pi}{2}}\frac{\sin \theta\;d\theta}{A^2 \cos^2 \theta+\epsilon^2}=\left[-\frac{2}{A}\arctan\left(\frac{A\cos(\theta)}{\epsilon}\right)\right]_0^{\pi/2}= \frac{2}{A}\arctan\left(\frac{A}{\epsilon}\right)$$and the right-hand side has a finite limit as \epsilon\to 0. Therefore$$\lim_{\epsilon\to 0}I(\epsilon)=\frac{2i}{A}... 1 They want you to integrate by parts. Try to see if that helps; I've added a spoiler for you below, if you need it. 1 I don't think $$P(\bigcup_{n \ge 1} \bigcap_{k \ge n} \{|X_k - X| < \epsilon\}) = 1, \text{ for all \epsilon > 0}$$ is the same as $$P(\bigcap_{m \ge 1} \bigcup_{n \ge 1} \bigcap_{k \ge n} \{|X_k - X| < 1/m\}) = 1.$$ 1 Break your integral in two:$$\int_0^1\frac{\mathrm dx}{(x^a+x^b)^p}+\int_1^\infty\frac{\mathrm dx}{(x^a+x^b)^p}.$$Since$$\lim_{x\to\infty}\frac{\frac1{x^{bp}}}{\frac1{(x^a+x^b)^p}}=1,$$the second integral converges if and only if the integral$$\int_1^\infty\frac{\mathrm dx}{x^{bp}}$$converges. And, since$$\lim_{x\to0}\frac{\frac1{x^{ap}}}{\frac1{(x^a+x^b)^p}}... 1 Let n >1. \log n = \displaystyle{ \int_{1}^{n}}\dfrac{1}{x}dx \lt \displaystyle{\int_{1}^{n}}\dfrac{1}{x^{1/2}}dx = 2x^{1/2} \big ]^n_1=2n^{1/2} -2; \dfrac{\log n}{3n^2+2}\lt \dfrac{2n^{1/2}-2}{3n^2+2}\lt \dfrac{2n^{1/2}}{3n^2}= (\dfrac{2}{3})\dfrac{1}{n^{3/2}}. Hence? 1 The general term is asymptotically equivalent to \;\dfrac{\log n}{3n^2}. Now, a Bertrand's series \;\displaystyle\sum_{n\ge 2}\frac 1{x^\alpha \log^\beta n} is known to converge if and only if \alpha>1 or \alpha =1, \beta >1. In the present case, without referring to Bertrand's series, you may split the general term as$$ \frac13\,\underbrace{... 1 Let$M>0$.$|X_n(Y_n-c)|> \epsilon$implies$|X_n| >M$or$|Y_n-c| >\frac {\epsilon} M$. Uniform integbrability implies that given$\eta >0$we can find$M$such that$P(|X_n| >M)<\eta$for all$n$. Hence$P(|X_n(Y_n-c)|> \epsilon) <\eta+ P(|Y_n-c| >\frac {\epsilon} M$. The second term tends to$0$so the proof is complete. 1 Unless$|\sin(x)|=1$, the series converges (absolutely) by the ratio test for any values of$\alpha$. This means that it converges absolutely for any values of$x$that are not in the form$2\pi k \pm \pi/2$. If$x$is in the form$2\pi k + \pi/2$, then$\sin(x)=1$and we are left with a typical p-series, which we know to converge for all$\alpha > 1\$. ...

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