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9

Hint. One may observe that $$ \sin^3 (a)=\frac34\sin (a)-\frac14 \sin(3a) $$ giving $$ 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\frac{3^{n}}4\sin\left(\frac{\pi}{3^{n+1}}\right)-\frac{3^{n-1}}4\sin\left(\frac{\pi}{3^{n}}\right) $$ then one gets a telescoping sum.


8

Hint $$\int\frac{1}{x\ln(x)}\mathrm d x=\ln(|\ln(x)|).$$


5

Using d'Alembert, $$\lim_{n\to \infty }\left|\frac{(n+1)+2}{n+1}\right|=1.$$ Therefore, the serie converge if $|x-1|<1$ and thus, for $x\in (0,2)$. Therefore, your statement is true.


5

We have that $\sin(x)-x\cos(x)$ behaves like $x^3$ in a right neighbourhood of the origin, hence integrability over there is ensured by $\color{red}{\alpha < 4}$. By Dirichlet's test, $$ \int_{1}^{+\infty}\frac{\sin x}{x^\beta}\,dx,\qquad \int_{1}^{+\infty}\frac{\cos x}{x^\beta}\,dx $$ are convergent as soon as $\color{red}{\beta>0}$, hence the ...


4

Use a ratio test :$$\lim _{ n\rightarrow \infty }{ \left| \frac { { a }_{ n+1 } }{ { a }_{ n } } \right| = } \lim _{ n\rightarrow \infty }{ \left| \frac { { \left( n+1 \right) }^{ 3 }{ e }^{ -\left( n+1 \right) } }{ { n }^{ 3 }{ e }^{ -n } } \right| =\lim _{ n\rightarrow \infty }{ \left| \frac { { \left( n+1 \right) }^{ 3 } }{ e{ n }^{ 3 } } \right| ...


3

Using the ratio test, $$\lim_{n\rightarrow\infty} \left|\frac{[(n+1)+2](x-1)^{n+1}}{(n+2)(x-1)^n}\right|=\lim_{n\rightarrow\infty}\left|\frac{(n+3)}{(n+2)}(x-1)\right|=|x-1|$$ The series converges if $|x-1|<1$. Thus, $x\in(0,2)$. However, we must check the boundaries: when $x=2$ and when $x=0$ as sometimes, depending on the series, it may be convergent....


3

First, rewrite the exponenent as \begin{align*} \exp \left( n\left( a\frac{S_n}{n} - b \right) \right) \end{align*} By the Law of Large Numbers, we have $a S_n/n- b \to a \mathbb{E}(\xi_1) - b$ a.s.. So $a S_n -bn \to - \infty$ a.s. if and ony if $b>a \mathbb{E}(\xi_1) =0$, and thus \begin{align*} \exp( a S_n - b n ) \to 0 \text{ iff } b > 0. \end{...


3

Hint. As suggested by @Claude Leibovici, one may write, as $k \to \infty$, $$ \left|\frac{\binom{2(k+1)}{k+1}x^{k+1}}{\binom{2k}{k}x^{k}}\right|=\frac{2(k+1)(2k+1)}{(k+1)^2}|x| \to 4|x|. $$ Then the radius of convergence is $R=\dfrac14$.


3

Hint. By the change of variable $x=t^{1/4}$, one gets $$ \int_{1}^{\infty} x^2 \sin(x^4) dx=\frac14\int_{1}^{\infty}\frac{\sin t}{t^{1/4}} dt $$ The latter integral is convergent by applying the Dirichlet test of convergence for improper integrals. Alternatively, one may integrate by parts, for $M\geq1$: $$ \int_{1}^M\frac{\sin t}{t^{1/4}} dt=\...


3

Directly ratio test: $$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|\alpha-1|(n+1)!}{(n+2)!-(n+1)!+1}\frac{(n+1)!-n!+1}{n!}=$$ $$=|\alpha-1|\frac{\left(n\cdot n!+1\right)(n+1)}{(n+1)(n+1)!+1}=|\alpha-1|\frac{n+\frac1{n!}}{(n+1)+\frac1{n!}}\xrightarrow[n\to\infty]{}|\alpha-1|$$ and the series converges exactly when $\;|\alpha-1|<1\;$


3

Why wouldn't we be able to use the ratio test? Let $b_n=a_nx^n$ The ratio test states if we have the series $\sum_{n=0}^\infty b_n$ and we define $L=\lim_{n\rightarrow \infty} \left|\dfrac{b_{n+1}}{b_n}\right|$, then $L$ converges absolutely if $L<1$, $L$ diverges if $L>1$, and $L$ is inconclusive if $L=1$. $$L=\lim_{n\rightarrow \infty}\left|\...


2

$$\sum_{n=0}^\infty 3^{n-1}\sin^3\left(\frac{\pi}{3^{n+1}}\right)$$ By $\sin3\theta =3\sin \theta-4\sin^3\theta$ $$4\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\sin\left(\frac{3\pi}{3^{n+1}}\right)-3\sin\left(\frac{\pi}{3^{n+1}}\right)$$ Then $$\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}\cdot4\sin^3\left(\frac{\pi}{3^{n+1}}\right)=\frac{1}{4}\sum_{n=0}^\infty 3^{n-1}...


2

We can show that for any integer $k$, the sequence $\left(\mathbb E\left[\left(\frac 1{\sqrt n}\sum_{i=1}^nX_i\right)^{2k}\right]\right)_{n\geqslant 1}$ is bounded, by expanding the sum. Therefore, we have the uniform integrability of the sequence $\left(f\left(n^{-1/2}\sum_{i=1}^n X_i\right)\right)_{n\geqslant 1}$ where $f(x)=\left|x\right|^p$ for any $p\...


2

hint: $$\sum_{n=1}^{\infty} \frac{n}{\sqrt{4+n^\alpha}}<\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^\alpha}}=\sum_{n=1}^{\infty} n^{1-\frac{\alpha}{2}}$$ now applying well known fact that $$\sum_{n=1}^{\infty}n^x$$ converges when $x<-1$ thus $$1-\frac{\alpha}{2}<-1$$ $$\alpha>4$$


2

Hint. One may observe that, as $n \to \infty$, we have $$ \frac{n}{\sqrt{4+n^\alpha}}=\frac1{n^{\alpha/2-1}}\frac1{\sqrt{1+4/n^\alpha}} \sim \frac1{n^{\alpha/2-1}} $$ which gives a convergent series if and only if $\alpha/2-1>1$ that is if and only if $\alpha>4$. Then this is the case for the initial series by using the comparison test.


2

It is convergent. To see this, use equivalents: $(n-1)^{1/2}\sim_\infty n^{1/2}$ $(n+1)^2-1\sim_\infty n^2$ Hence $\;\dfrac{(n-1)^{1/2}}{(n+1)^2-1}\sim_\infty\dfrac{n^{1/2}}{n^2}=\dfrac1{n^{3/2}}$, which converges.


2

For $a\ne -1$, we have using integration by parts $$\begin{align} I(a)&=\lim_{\epsilon \to 0^+}\int_\epsilon^1 x^a \log(x)\,dx\\\\ &=\lim_{\epsilon \to 0^+}\left(\left.\frac{x^{a+1}\log(x)}{a+1}\right|_{\epsilon}^1-\frac{1}{a+1}\int_\epsilon^1 x^{a}\,dx\right)\\\\ &=\lim_{\epsilon \to 0^+}\left(-\frac{\epsilon^{a+1}\log(\epsilon)}{a+1}-\frac{1}{(...


2

Suppose, for the moment, that the family $\{f_n\}_n \cup \{f\} \subseteq L^1$ is uniformly integrable, i.e. that for any $\epsilon>0$ there exists $\delta>0$ such that $$\sup_{n \in \mathbb{N}} \int_A |f_n| \, d\mu + \int_A |f| \, d\mu < \epsilon \tag{1}$$ for any Borel set $A$ with $\mu(A)\leq\delta$ (here, and in what follows, $\mu$ denotes a ...


2

Try with $f(n):= \frac{1}{n \log(n)^2}$ and the Bertrand sum criterion.


1

Your approach can be continued. As mentioned in the comments $n^{3/n} \to 1$ as $n \to \infty$. Note that this follows if we can prove $n^{1/n} \to 1$ as $n \to \infty$. Here is an elementary proof using only the (generalized) squeeze theorem and basic algebra. Take any real $ε > 0$, and let $c \in \mathbb{N}$ such that $1/2^c \le ε$.   [This arises ...


1

If an infinite sum converges this implies that its terms $a_n <1 \ \forall \ n >N$, and for values $|a_n|<1$ clearly $a^2_n < a_n$.


1

Define a sequence $\{F_n\}$ for $n\ge1$ by the equation \begin{equation} \int_{F_n}^{1}\sqrt{1-x^2}\,dx=\frac{\pi}{n} \end{equation} Then clearly $\{F_n\}$ in an increasing sequence approaching $1$ as a limit. Therefore the series $\sum_n^\infty F_n$ and $\sum_{n=1}^\infty (-1)^{n}F_n$ both diverge since $\lim_{n\to\infty}F_n\ne0$. However, consider the ...


1

Hint: write it as follows you will see $$\int _{ 0 }^{ 3 } \: \frac { x }{ \left( 9-x^{ 2 } \right) ^{ \alpha } } dx=\frac { 1 }{ 2 } \int _{ 0 }^{ 3 } \: \frac { d\left( 9-{ x }^{ 2 } \right) }{ \left( 9-x^{ 2 } \right) ^{ \alpha } } ={ \frac { \left( 9-x^{ 2 } \right) ^{ -\alpha +1 } }{ -2\alpha +2 } }_{ 0 }^{ 3 }=-\frac { 9^{ -\alpha +1 } }{ -2\alpha ...


1

It's a power series with center in $x=1$; you can calculate the radius of convergence: $$R=\dfrac{1}{\limsup_{n\to +\infty}\sqrt[n]{\lvert a_n\rvert}}$$ where $a_n=n+2$ You have: $$\limsup_{n\to +\infty}\sqrt[n]{\lvert a_n\rvert}=\limsup_{n\to +\infty}\sqrt[n]{n+2}=1$$ so $R=1$ The disc of convergence is: $$\lbrace x\in\mathbb{R} | \lvert x-1\rvert<R\...


1

$\int _0^{\frac{1}{2}}\frac{1}{x\ln \left(x\right)}dx=-\infty \:$ $\mathrm{Apply\:Integral\:Substitution:}\:\int f\left(g\left(x\right)\right)\cdot g^{'}\left(x\right)dx=\int f\left(u\right)du,\:\quad u=g\left(x\right)$ $u=\ln \left(x\right)\quad \:du=\frac{1}{x}dx$ $=\int \frac{1}{u}du$ $\mathrm{Substitute:}\:u=\ln \left(x\right)$ $\frac{...


1

Hint. One may write, as $n \to \infty$, $$ \frac{n!}{\left(n+1\right)!-n!+1} =\frac1{n+1/n!} \sim \frac1n $$ and the initial series is convergent for $|\alpha-1|<1$. When $|\alpha-1|=1$ we obtain a conditionally convergent series.


1

I won't insert all the details into a hint, but consider this oft-used technique: $|x(t_2)-x(t_1)|<|x(t_2)-x_m(t_2)|+|x_m(t_2)-x_m(t_1)|+|x_m(t_1)-x(t_1)|$, the RHS of which is arbitrarily small. The existence of the function $x(t)$, the 'limit' of the Cauchy sequence, or more formally that function for which $\sup_{0\leq t\leq T}|x(t)-x_m(t)|$ is ...


1

If one knows the following Taylor series expansion, as $u \to 0$, $$ \log(1+u)=u+O(u^2) $$ then one may write, for any fixed real number $t$, as $n \to \infty$, $$ \left( 1+\frac{t}{n} \right)^{n}=e^{\large n\log\left(1+\frac{t}{n}\right)}=e^{\large n\left(\frac{t}{n}+O\left(\frac{1}{n^2}\right)\right)}=e^{t+O\left(\frac{1}{n}\right)} $$ which gives $$ \...



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