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4

If $f_n,f\geq 0$ then you can consider the sequence $$g_n=\min(f_n,f)=\frac{1}{2}(f_n+f-|f_n-f|)$$ which is bounded by $f$ and and converges pointwise a.e. to $f$. Then use the dominated convergence to obtain the claim. Generally, if you allow $f_n,f$ to attain values in $\mathbb R$, the claim is not true. Consider the functions $f_n= n ...


4

You may use the Maclaurin expansion for $\cos x$: $$ \cos (x) = 1 -\frac{x^2}{2}+ \mathcal{O}(x^4) $$ giving $$ \sum_{n=2}^{\infty}\left( 1-\cos \left(\frac{\pi}{n}\right)\right)=\sum_{n=2}^{\infty}\frac{\pi^2}{2n^2}+\sum_{n=2}^{\infty}\mathcal{O}(\frac{1}{n^4}) $$ concluding that your initial series converges.


4

Hint From an algebraic point of view, consider $$S=\sum_{n=1}^{\infty}(2n-1)x^n$$ and you can write $$S=2x\sum_{n=1}^{\infty}n x^{n-1}-\sum_{n=1}^{\infty} x^{n}$$ I am sure that you can take from here. When you finish, replace $x$ by $\frac12$.


4

By the Taylor series we have $$\left(1+\frac1n\right)^n=\exp\left(1-\frac1{2n}+\mathcal O\left(\frac1{n^2}\right)\right)=e-\frac e{2n}+\mathcal O\left(\frac1{n^2}\right)$$ so we see that this series is convergent since $\sum\frac{(-1)^n}{n}$ is convergent by Leibniz theorem and the series $\sum O\left(\frac1{n^2}\right)$ is convergent by comparison with the ...


4

Cauchy-Schwarz says $$ \sum_{n=1}^\infty\frac{|a_n|}{n}\le\left(\sum_{n=1}^\infty a_n^2\right)^{1/2}\left(\sum_{n=1}^\infty\frac1{n^2}\right)^{1/2} $$ so the series is absolutely convergent.


3

Keyword: Telescoping series. For every $n\geqslant1$, let $A_n=a_1+\cdots+a_n$, then, for every $n\geqslant2$, $A_n\geqslant A_{n-1}$ hence $$\frac{a_n}{A_n^2}\leqslant\frac{a_n}{A_nA_{n-1}}=\frac1{A_{n-1}}-\frac1{A_n}.$$ Summing these yields, for every $N$, ...


3

Here is a sledgehammer solution: Using the Hardy's inequality, we get $$ \sum_{n=1}^{\infty} \frac{a_{n}^{2}}{n^{2}} < \infty. $$ Now you can apply the Kronecker's lemma to obtain the desired conclusion. This at least shows that your guess is correct. I believe that there is a much simpler solution, and I will update my solution when I find it.


3

The limits are unique - more precisely, they are unique in distribution, i.e. if $X_n \to Y$ in distribution and $X_n \to Z$ in distribution, then $Y$ and $Z$ have the same distribution. If you define a metric $d$ such that $$d(X_n,Y) \to 0$$ if and only if $X_n \to Y$ in distribution, then you actually define a metric on the space of distributions. This ...


3

Hint: Our denominator is $\exp((\ln\ln n)(\ln n))$, which is $n^{\ln\ln n}$. Now think $p$-series and Comparison. Remark: When we meet $a^b$, the fact that it is equal to $\exp((\ln a)(b))$ is frequently useful.


3

This sequence does not tend to infinity and it is not unbounded. $$ \lim_{n\to\infty}a_n=3. $$ Proof. First we prove that the sequence is decreasing. Clearly, $0<a_2=\sqrt{2\cdots 4+3}<4=a_1$. Inductively, if $a_k<a_{k-1}$, then $\,0<a_{k+1}=\sqrt{2a_{k}+3}<\sqrt{2a_{k-1}+3}=a_{k}$. Hence, $\{a_n\}_{n\in\mathbb N}$ is decreasing and lower ...


3

$$S:=\sum_{n=1}^\infty\frac{(H_n)^3}{n^3}$$ is convergent since the terms are equivalent to $\,\left(\dfrac{\ln(n)}n\right)^3\,$ as $\;n\to \infty\,$ and given by $$S=\frac {31}{5040}\pi^6-\frac 52\zeta(3)^2\\\approx 2.3009545517005250398$$ I don't know a complete proof but we may start with the following class of Euler sums : ...


2

$\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}<\frac{1}{1*2}+\frac{1}{2*3}+...+\frac{1}{(n-1)*n}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}<1 $, so $ x_n<2 $ for all $ n $.


2

Using the properties of logarithms, I write $\ln(n^2/(n^2-1))$ as $2\ln(n)-\ln(n-1)-\ln(n+1)$ and then I consider the partial sum $S_n= (2\ln(2)-\ln(1)-\ln(3))+(2\ln(3)-\ln(2)-\ln(4))+(2\ln(4)-\ln(3)-\ln(5))+...+(2\ln(n)-\ln(n-1)-\ln(n+1))$ Which after canceling like terms simplifies to $S_n=\ln2+\ln(n)-\ln(n+1)=\ln(2n/(n+1))=\ln(2/(1+(1/n)))$ And ...


2

I should start with 2) using induction. 2) If $1<a_n\leq 2$ then: $$a_{n+1}=\frac{a_n+1}{2}>\frac{1+1}{2}=1$$ and $$a_{n+1}=\frac{a_n+1}{2}\leq\frac{2+1}{2}\leq 2$$ 1) From $a_n>1$ it follows directly that: $$a_{n+1}=\frac{a_n+1}{2}<a_n$$ A bounded monotone sequence is convergent and we can find its limit $a$ on base of the relation ...


2

You have made a little mistake. Correction: $$ A_1 = \lim_{n \rightarrow \infty} (A_1^n)^{1/n} \leq \lim_{n \rightarrow \infty}(A_1^n + ... A_k^n)^{1/n} \leq \lim_{n \rightarrow \infty} (kA_1^n)^{1/n} = \lim_{n \rightarrow \infty} A_1{\color{blue}{k^{1/n}}}=A_1 $$


2

Observe that: $\displaystyle \lim_{n \to \infty} \dfrac{\ln\left(1+\dfrac{1}{n+1}\right)}{\dfrac{1}{n+1}} = 1$, and the harmonic series $\displaystyle \sum_{n=1}^\infty \dfrac{1}{n+1}$ diverges, so the series $\displaystyle \sum_{n=1}^\infty \ln\left(1+\dfrac{1}{n+1}\right)$ also diverges.


2

Hint: $$\sum \frac{1}{\sqrt{n+3/4}}= \sum \frac{2}{\sqrt{4n+3}} \leq \sum \frac{1}{\sqrt{4n+1}}+ \frac{1}{\sqrt{4n+3}} \leq \sum \frac{2}{\sqrt{4n}}=\sum \frac{1}{\sqrt{n}}.$$ Try cases $a=\sqrt{2}$ and $a\neq \sqrt{2}$, and check if they ever converge...


2

Here is a solution using Cauchy's inequality and Stolz–Cesàro theorem. By Stolz–Cesàro theorem, we need to prove $$\lim\dfrac{1}{n^2}\sum_{k=1}^n a_k^2 = \lim \dfrac{a^2_{n+1}}{2n+1} =0$$ provided the second limit exists. Let $b_k = a_{k+1}- a_k$ with $b_0 = a_1$, we have $\sum_k b_k^2 < \infty$ and $a_{n+1} = \sum_{k=0}^{n}b_k$. \begin{align} ...


2

Hint: $$ \sqrt {AB}\le A+B $$ because of $(\sqrt A - \sqrt B)^2 \ge 0$.


2

For two: First you can prove by induction that $a_n > 1$ Looking for limit point, they have to verify : $L^2 - L -1 = 0$ which gives you only one value >1 (noted g here) Now, let's prove that $a_n \leq g$ by induction: $a_1 = 1 < g$ ; if $a_n \leq g$ : $a_{n+1} = \sqrt{1 + a_n} \leq \sqrt{1 + g} = g $ Now there is a lemma that says : If $a_n$ ...


2

As has been mentioned, $(1+n^2)^{-1/4}\ge\frac1{\sqrt{2n}}$ which diverges by the integral test. Since $\cos(x+\pi)=-\cos(x)$, we have that $$ \begin{align} \sum_{k=n}^{n+11}\cos\left(\frac{k\pi}6\right) &=\sum_{k=n}^{n+5}\left[\cos\left(\frac{k\pi}6\right)+\cos\left(\frac{k\pi}6+\pi\right)\right]\\ ...


2

$$\sum_{n=1}^\infty\frac{2n-1}{2^n}=\sum_{n=1}^\infty\frac n{2^{n-1}}-\sum_{n=1}^\infty\frac1{2^n}=\frac1{\left(\frac12\right)^2}-\frac{\frac12}{1-\frac12}=4-1=3$$ The above follows from $$|x|<1\implies\frac1{1-x}=\sum_{n=0}^\infty x^n\;,\;\;\left(\frac1{1-x}\right)'=\sum_{n=1}^\infty nx^{n-1}$$ and the splitting of the first sum in the first line ...


2

What you write is the probability of $P(\sum_{k=1}^nX_n \geq \lfloor \frac{n}{2}\rfloor)$, where $X_i$'s are i.i.d Bernoulli variable with parameter $Q$. $P(\sum_{k=1}^nX_n \geq \lfloor \frac{n}{2}\rfloor) = P(\dfrac{\sum_{k=1}^nX_n -nQ}{\sqrt{n}} \geq \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}})$ and \begin{align} \lim_{n\to \infty} \dfrac{\lfloor ...


2

Hint: if $x_n\to x$ then $x_{n+1}-x_n \to 0$. But...


2

On one hand, the series is strictly increasing. On the other hand, $n^3+2\le2n^3$ and $n^4+3n^2+1$ $\ge n^4$, implying $t_n\le\ldots~\Big($I'll let you fill in the dots, and draw the conclusion for yourself$\Big)$.


2

Let $a_{n} = \frac{\sqrt{n^{3}+2}}{n^{4}+3n^{2} + 1}$ and let $b_{n} = \frac{1}{n^{\alpha}}$ for some positive $\alpha$. Consider $$\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}} = \lim_{n\rightarrow \infty} \frac{\frac{\sqrt{n^{3}+2}}{n^{4}+3n^{2} + 1}}{\frac{1}{n^{\alpha}}} = \lim_{n\rightarrow \infty} \frac{n^{\alpha}\sqrt{n^{3}+2}}{n^{4}+3n^{2} + 1} = ...


2

Hint: Fix $a\in(0,1)$, Then $$(-1)^n\sin \frac{a}{n} = (-1)^n\left(\frac{a}{n} - \frac{a^3}{6n^3} + o\!\left(\frac{1}{n^3}\right)\right) = a_n + b_n$$ where $a_n=\frac{(-1)^n a}{n}$ and $b_n = \frac{(-1)^n a^3}{6n^3} + o\!\left(\frac{1}{n^3}\right)$. Now, $\sum a_n$ is conditionally convergent, $\sum b_n$ absolutely convergent; what does that imply for the ...


2

$\sin(\frac{a}{n})>0$ and decreases to 0, hence by alternating test, the series converges. Since $|\sin(x)|>\frac{2x}{\pi}$ for $|x|<\frac{\pi}{2}$, so $|\sin(\frac{a}{n})|>\frac{2a}{\pi n}$, hence by comparison test, it's not absolutely convergent.


2

Hint: Use that $(1+\epsilon)^n\ge1+n\epsilon$ by Bernoulli's inequality, and $1+n\epsilon>3 \iff n\epsilon>2\iff n>\frac{2}{\epsilon}$.


1

If you know this theorem, it will be helpful for you. If $\lim_{n\to \infty}{x_n}=A$ and $\lim_{n\to \infty}{y_n}=B$, then $\lim_{n\to \infty}{x_n}^{y_n}=A^B$



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