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16

You could try to prove the general, useful fact that $$\lim_{n\to\infty}a_n=L\implies\lim_{n\to\infty}\frac{a_1+\ldots+a_n}n=L$$


7

We have by the Riemann sum $$c_n=\frac1n\sum_{k=2}^n\frac1{\sqrt k}=\frac1{n^{3/2}}\sum_{k=2}^n\frac1{\sqrt{\frac kn}}\sim\frac1{\sqrt n}\int_0^1\frac{dx}{\sqrt x}=\frac2{\sqrt n}$$ so clearly $c_n$ tends to $0$ but we found also that $$\lim_{n\to\infty}c_n\sqrt n=2$$ which we can't find it by applying the Cesàro's theorem.


5

Hint : $c_n$ is Cesàro summation of sequence $\{\frac{1}{\sqrt{n}}\}_{n \in \Bbb N}$ Cesàro summation


5

Let $E_1 = [0, 1/2]$; $E_2 = [1/2, 1]$ $E_3 = [0,1/4]$; $E_4 = [1/4,1/2]$; $E_5 = [1/2,3/4]$; $E_6 = [3/4,1]$ $E_7 = [0,1/8]$; $E_8 = [1/8,1/4]$; etc. Then $\mu(E_n) \rightarrow 0$, but for any $x \in [0,1]$, it's easy to see that $f_n(x) = f(x)\chi_{E_n^c}(x)$ differs from $f(x)$ for infinitely many $n$, so $f_n(x)$ does not converge pointwise to $f(x)$ ...


5

Base of induction Since $a_1=1$ and $a_2=2/3$, the inequality $a_{n+1}\le a_n$ is true for $n=1$. Inductive step If $a_{n+1}\le a_n$, then $a_{n+1}^2\le a_n^2$, since all $a_n$ are positive by definition. Hence $$a_{n+2} = \frac13\left(a_{n+1}^2+\frac1{n+1}\right)\le \frac13\left(a_n^2+\frac1n\right) = a_{n+1}$$ which establishes the inductive step. ...


5

The maximum $M_n$ of interest is such that $$(1-2/\sqrt{n})K_n\leqslant M_n\leqslant1,\quad\text{where}\quad K_n=\max\{X_i\mid 1\leqslant i\leqslant\sqrt{n}\}.$$ By independence, $P(K_n\leqslant x)=P(X_1\leqslant x)^{\sqrt{n}}=x^{\sqrt{n}}\to0$ when $n\to\infty$, for every $x$ in $(0,1)$. Hence $K_n\to1$ in probability, $(1-2/\sqrt{n})K_n\to1$ in ...


4

The given series is the zeta Riemann series and it's pointwise convergent on $(1,+\infty)$. Moreover for all $a>1$ we have $$\frac{1}{n^x}\le \frac1{n^a},\quad \forall x\ge a$$ and since $\sum\frac1{n^a}$ is convergent then we have uniform convergence on every interval $[a,\infty)$. The given series isn't uniformly convergent on $(1,\infty)$. Indeed ...


4

We have $c_n>0$ and $$\begin{align}c_{n+1}-c_n&=\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}-\cdots-\frac{1}{\sqrt{n}}\right]\\ &<\frac{1}{n(n+1)}\left[\frac{n}{\sqrt{n+1}}-\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n}}-\cdots-\frac{1}{\sqrt{n}}\right]\\ ...


3

Use the Dominated convergence theorem. $ f_ n$ converges pointwise to $0$ and is bounded by $1/\sqrt x$ which is integrable.


3

We note that the argument of the $\ln$ is just a partial sum of the series. We know that if the series converges, the terms must converge to zero, so the argument of the $\ln$ must go to to $1$, so the sum must go to $1$. To establish convergence, we are given $x_1 =\sum_{n=1}^1 x_n \gt 1$ as the base case of an induction. Assume $\sum_{n=1}^k x_n \gt ...


3

You are probably assuming too much. If $a_n \to a$ and $b_n \to b$, then $-a_n b_n \to -ab$. Thus $$\limsup_{n \to \infty} (-a_nb_n) = \lim_{n \to \infty}(-a_nb_n) = -ab.$$


3

I have some partial results; perhaps someone can continue/correct my work. \begin{align*} &\phantom{{}={}}P\left(1-\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}>\epsilon\right)\\ &=P\left(1-\epsilon>\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}\right)\\ &=\prod_{i=1}^{n/2} P\left(1-\epsilon ...


3

You are correct that the series is convergent, for the reasons you indicated. The answer key is checking for absolute convergence. The series is convergent, but not absolutely convergent. In other words, it is conditionally convergent.


2

A necessary condition for a series to converge is that $$\lim_{n \to \infty} a_n = 0$$ This has nothing to do with the alternating series test. If one of the other hypothesis fails, then one cannot conclude divergence.


2

Hint for the convergence in $L^1$ (since the almost sure convergence seems dealt with in the question): For every $n$, $$E\left(\sum_{k=n+1}^\infty X_k\right)=\frac{\mathrm e^{-n}}{\mathrm e-1},$$ and the RHS converges to zero when $n\to\infty$.


2

First we see that a necessary condition to have the limit $e^\alpha$ is that $\lim\limits_{n\to \infty}a_n=0$. Second using Taylor expansion we get $$\left(1+\alpha a_n\right)^n=\exp(n\ln(1+\alpha a_n))\sim_\infty\exp(\alpha na_n)\xrightarrow{n\to\infty}e^\alpha\iff \lim_{n\to\infty}na_n=1\iff a_n\sim_\infty\frac1n$$


2

Assume that $X=\{x_j,j\in\mathbb N\}$ and that $\mu\{x_j\}>0$ for each $j$ and $f=0$, $f_n\geqslant 0$. We have to show that for each $j$, $f_n(x_j)\to 0$ as $n$ goes to infinity. Fix $j$ and $\varepsilon\gt 0$; for $n$ large enough, we have $\mu\{x,f_n(x)\geqslant\varepsilon\}\lt\mu\{x_j\}$, hence $f_n(x_j)\lt\varepsilon$ for these $n$. In general, ...


2

You should have gotten $\displaystyle\sum_{n = 1}^{\infty}\dfrac{(-1)^n}{n} = \sum_{n = 1}^{\infty}(-1)^n(I_{n-1}+I_n)$. (You forgot the summation sign). Now, look at the partial sums: $\displaystyle\sum_{n = 1}^{N}(-1)^n(I_{n-1}+I_n) = -(I_1+I_2)+(I_2+I_3)-(I_3+I_4)+\cdots+(-1)^N(I_{N-1}+I_{N})$. A lot of terms cancel. Now, take the limit as $N \to ...


2

Hint. Let $\delta>0$, then, for $x \in [1+\delta,+\infty)$, $$ \frac{1}{n^x}\leq \frac{1}{n^{1+\delta}},\quad n=1,2,\ldots $$ Since the series $\displaystyle \sum_{n\ge 1}\frac {1}{n^{1+\delta}}$ is convergent (compare it to the integral $\displaystyle \int_1^{+\infty}\frac {dx}{x^{1+\delta}}$), then $\displaystyle \sum_{n\ge 1}\frac {1}{n^x}$ is ...


2

We know that if $n_k$ is a sub sequence of $s_n$, then $k \ge n$. By convergence, there exists $N$ so that $|s_n - A| < \epsilon$ whenever $n \ge N$. But, we have $n_k \ge n$ so that $$n_k \ge n \ge N.$$ Hence, $$|s_{n_k} - A| < \epsilon$$ whenever $n_k \ge N$. Hence, $s_{n_k} \longrightarrow A$. In response to the comments below: Consider the ...


2

The idea here is that you choose a pair of numbers, $n$ and $m$ such that $n\geq N$ for the appropriate $N$, implying you have both that $$|x_n-x|<\varepsilon'/2$$ $$|x_m-x|<\varepsilon'/2.$$ From which point the proof follows by the triangle equality you set up. We're not saying $|x_n-x|<\varepsilon'/2$ because $n$ is special - we're saying that's ...


2

Note that $\forall \epsilon>0,\ \forall n\in \mathbb{N},\ \exists N\in \mathbb{N}$ such that $\forall n\ge N$$$|s_n-\sigma|\le |s_n-\sigma_n|+|\sigma_n-\sigma|\le \left|\frac{\sum_{k=1}^n ka_k}{n+1}\right|+\epsilon/2\le \frac{n}{n+1}M+\epsilon/2\ \le \frac{N}{N+1}M+\epsilon/2\le \epsilon/2+\epsilon/2=\epsilon$$ where $N$ is chosen accordingly. So, by ...


2

Use direct comparison in both cases. In the first case: $$\sum_{k=1}^{\infty}\frac{k}{10+k^{2}}=\sum_{k=1}^{\infty}\frac{1}{10/k+k}\ge \sum_{k=1}^{\infty}\frac{1}{10+k},$$ from where the series is divergent. In the second case: $$0\le \sum_{k=1}^{\infty}\frac{1.3.5..(2k+1)}{4^{k} k!}\le \sum_{k=1}^{\infty}\frac{3^k}{4^{k}},$$ from where it follows that ...


2

Your bracketing $0 < c_n < \frac{1}{\sqrt{2}}$ shows that the sequence is bounded, but is not enough to prove convergence. Actually, bounding all terms by $1/\sqrt2$ will not lead you to the solution because it is too loose: thanks to the $1/\sqrt n$ decrease, most of the terms are much smaller than that. Now try to bound above as follows: ...


2

There is a tail of the sequence such that the terms in the tail are close enough to a certain term in the sequence. Now we just have to take care of the terms in the sequence that are not in that tail, and this is simple because there are only finitely many such terms.


2

The fact that $0 \not\in \mathbb{N}$ (if you use that convention) is not relevant. What you need to do is choose an $\varepsilon > 0$ such that $|x_n - x| < \varepsilon$ implies that $x_n = x$. In general you can't do this, but in this situation you can. As a hint for which $\varepsilon$ to choose, what is the smallest $|x_n - x|$ can be without ...


2

It's much easier than you think! Since by hypothesis we know that the conditional mean is $ E[X_i \mid Y_i] = Y_i $ and $Y_i$ is uniform with mean zero, by using the tower property we find: $$ n\cdot E [ \bar{X}_n ] = E\left[\sum_{i=1}^n X_i \right] = \sum_{i=1}^n E[X_i] = \sum_{i=1}^n E[E[X_i \mid Y_i]] = \sum_{i=1}^n E[Y_i] = 0 $$


1

$$A=2\pi\int_1^\infty x^p \sqrt{1+p^2x^{2p-2}}dx=2\pi\int_1^\infty \sqrt{x^{2p}+p^2x^{4p-2}}dx$$ If $p>1$, then $x^{4p-2}\ge x^{2p}$ so $$\int_1^\infty \sqrt{x^{2p}+p^2x^{4p-2}}dx\geq \int_1^\infty \sqrt{x^{2p}+p^2x^{2p}}dx=\sqrt{1+p^2}\int_1^\infty x^p\,dx$$ which diverges. If $0<p<1$, then $x^{2p}\geq x^{4p-2}$ so \begin{align} \int_1^\infty ...


1

a) : Let $a_n = \frac{(-1)^n}{n^{\frac{1}{8}}}$. $\sum a_n$ converges from Abel's Test, if for nothing else. But, $b_n = a_n^4 = \frac{1}{\sqrt{n}}$ is not the general term of a converging series, since $\sum \frac{1}{n^a}$ converges iif a>1. b) : $\sum |a_n|$ converges $\implies (|a_n|) \rightarrow 0 \implies |a_n| < 1$ for $n \geq N_o$. So : $ n \geq ...



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