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5

Write $\frac 1r = a + 1$ where $a > 0$. If $n \ge 2$ you have $$\frac 1{r^n} = (a + 1)^n \ge \frac{n(n-1)}{2} a^2$$ according to the binomial theorem. Consequently $$0 \le n r^n \le \frac{2}{(n-1)a^2}$$ for all $n \ge 2$. Now let $n \to \infty$ and use the squeeze theorem.


4

We have, taking log and using L'Hopital, $$\lim_{n\rightarrow\infty}-t\sqrt{n}-n\log\left(1-\frac{t}{\sqrt{n}}\right)=\lim_{n\rightarrow\infty}\frac{\left(-\frac{t}{\sqrt{n}}-\log\left(1-\frac{t}{\sqrt{n}}\right)\right)}{1/n}=\frac{1}{2}t^{2}\lim_{n\rightarrow\infty}\frac{\sqrt{n}}{\sqrt{n}-t} $$ and so your limit.


3

No, there is no general formula yet to know the sum of $$S(p)=\sum_{n=1}^{\infty} \frac{1}{n^p}$$ But we know that it converges iff $p>1$ For every $p$ even we know that: $$S(p)=(-1)^{\frac p2+1}{{B_p(2\pi)^p}\over {2p!}}$$ where $B_n$ is a Bernoulli number. However for odd integers we still don't have a general formula, if you are interested in ...


3

The case $r=0$ is quite trivial, hence we assume $r\in(0,1)$. Let $s=-\log r\in\mathbb{R}^+$. We have to prove: $$ \lim_{n\to +\infty} n\cdot e^{-sn}=0, \tag{1}$$ that follows from: $$ 0\leq n\cdot e^{-sn} = \frac{n}{\left(e^{\frac{s}{2}n}\right)^2}\leq\frac{n}{\left(1+\frac{s}{2}n\right)^2}\leq\frac{4}{s^2 n}.\tag{2}$$


3

We have $a_n = -\log(\cos(1/n)) = -\frac12 \log(\cos^{2}(1/n)) = -\frac12 \log(1-\sin^{2}(1/n))$. We have $$\lim_{n \to \infty} \dfrac{a_n}{1/n^2} = -\frac12 \cdot \lim_{n \to \infty} n^2 \log(1-\sin^2(1/n)) = \frac12$$ Hence, by limit comparison test the series converges since $\sum_n \frac1{n^2}$ converges.


3

Find $N> -\log_{10}(3\epsilon)$. Then if $n>N$, then $10^n>\frac{1}{3\epsilon}$ or $\frac{1}{3}\cdot 10^{-n}<\epsilon$.


2

The marginal distribution of $x_i$ is binomial with parameters $n$ and $p_i$, so $E[x_i] = n p_i$, and thus $E[x_i - x_j] = n (p_i - p_j)$. The covariance matrix of $x_i$ and $x_j$ is $\pmatrix{n p_i (1-p_i) & -n p_i p_j\cr -n p_i p_j & n p_j (1-p_j)}$, so the variance of $x_i - x_j$ is $$\text{var}(x_i) - 2 \text{cov}(x_i, x_j) + \text{var}(x_j) = ...


2

Let $f$ be any function at all, say your favourite discontinuous function. Let $f_n=f$ for all $n$. Then the sequence $(f_n)$ converges uniformly. Somewhat less trivially, let $(g_n)$ be any uniformly convergent sequence of continuous functions and let $f_n=f+g_n$. Then the sequence $(f_n)$ converges uniformly, and the $f_n$ are not continuous if $f$ is not ...


2

$$(1)\;\;\;\sum_{n=1}^\infty\frac1n$$ $$(2)\;\;\;\sum_{n=1}^\infty\frac1{n^2}$$


2

Thomas Andrews is right when he says that there are no general rules. Often this type of problem, we must use a more intuitive method. I know the p-series for p = 2. The resolution method in this series comes from Euler himself. (1) $\sin(x) = x - x^3/3! + x^5/5! -x^7/7! + ...$ (Taylor series) (2) $\sin(x)/x = 1 - x^2/3! + x^4/5! -x^6/7! + ...$ (divided ...


2

from $ \frac {a_n} {a_{n-1}} \le \frac {b_n} {b_{n-1}} $ you get $$ a_N = \frac {a_0} {b_0} \times b_0 \times \prod_{n = 1}^{N} \frac {a_{n}} {a_{n-1}} \le \frac {a_0} {b_0} \times b_0 \times \prod_{n = 1}^{N} \frac {b_{n}} {b_{n-1}} = \frac {a_0} {b_0} \times b_N $$ so $$\sum b_N<\infty \implies \sum a_N<\infty $$


1

This follows because $f(x)=x^a$ is continuous, ie, $f(x_n) \to f(x)$ if $x_n \to x$. Edit: if you want to use the $\epsilon,\delta$ definition, observe that the binomial theorem gives $$x+h \leq (\sqrt[n]{x}+\sqrt[n]{h})^n $$ $$ \sqrt[n]{x+h} \leq \sqrt[n]{x}+\sqrt[n]{h}$$ $$\sqrt[n]{x+h} - \sqrt[n]{x} \leq \sqrt[n]{h}, (*)$$ for $x,h>0$. By picking ...


1

For the first part, you have established that $-5$ and $-3$ are not part of the radius of convergence, therefore the interval is $(-5,\,3)$, and not $[-5,\,3]$ (which would include $-5$ and $-3$). Same logic for the second part. You've shown that the series converges for $-5 < x < -3$ but diverges for $-5$ and $-3$, therefore, it is the open interval ...


1

This is a famous sum & there are multiple ways of approaching this. Here is one way via a geometric sum: We have the well known sum \begin{equation*} 1+x+x^2+\cdots =\frac{1}{1-x}. \end{equation*} Setting $x=e^{i\theta},~0<\theta<2\pi~(x\neq 1)$ gives \begin{equation*} 1+e^{i\theta}+e^{2i\theta}+\cdots ...


1

It is a Fresnel integral. The limit exists since $$ I(b)=\int_{0}^{b}\sin t^2\,dt = \frac{1}{2}\int_{0}^{b^2}\frac{\sin x}{\sqrt{x}}\,dx$$ converges by Dirichlet's test (integral version), because $\sin x$ is a function with a bounded primitive and $\frac{1}{\sqrt{x}}$ is a monotonic function converging to zero as $x\to +\infty$. To compute it, we may use ...


1

Suppose that $X_n\to X$ almost surely as $n\to\infty$ and $X_n\to Y$ almost surely as $n\to\infty$. Then there exists $\Omega'\subset\Omega$ such that $\Pr(\Omega')=1$ and for each $\omega\in\Omega'$ $$ |X_n(\omega)-X(\omega)|\to0 $$ as $n\to\infty$. Similarly, there exists $\Omega''\subset\Omega$ such that $\Pr(\Omega'')=1$ and for each $\omega\in\Omega''$ ...


1

I like this problem! I will have to assign this sometime. Apply the root test. The denominator satisfies $\lim_{n \to \infty} \sqrt[n]{2n+1} = 1$ (use L'Hopitals). The limit in the numerator simplifies to $\lim_{n \to \infty} |x-2|^{n} ,$ which is 0 when $|x-2|<1.$ Therefore the preliminary interval of converge is $1<x<3.$ The series is ...


1

You need to use the fact that if a sequence converges, any subsequence converges to the same limit. Since $\{y_{3n}\}$ converges to say $y$, then $\{y_{6n}\}$ must also converge to $y$. But $\{y_{6n}\}$ is a subsequence of $\{y_{2n}\}$ and since that sequence converges, it must converges to $y$. A similar argument will show that $\{y_{2n+1}\}$ will ...


1

Set $$S_n := \sum_{j=1}^n X_j.$$ Then $$\begin{align*} \frac{1}{n-1} \sum_{i=1}^n \left( X_i - \frac{X_1+\ldots+X_n}{n} \right)^2 &= \frac{1}{n-1} \sum_{i=1}^n \left( X_i - \frac{S_n}{n} \right)^2 \\ &= \frac{1}{n-1} \sum_{i=1}^n X_i^2 - 2 \frac{1}{n-1} \frac{S_n}{n} \sum_{i=1}^n X_i + \frac{1}{n-1} \frac{S_n^2}{n} \\ &= \frac{1}{n-1} ...


1

Your proof is correct. In fact, once you have $$|b_n|<\frac{M+2}{n}$$ it is clear that since the numerator is bounded one can make the right hand side $<\epsilon$ for arbitrarily small $\epsilon >0.$


1

Hint. The integrand is a continuous function on $(0,\infty)$, thus the potential problems are near $0$ and near $\infty$. As $x \to 0^+$, you have, for $b$ sufficiently near $0^+$: $$ \int_0^b \frac{1}{e^{\sqrt{x}}-1}dx \sim \int_0^b \frac{1}{\sqrt{x}}dx $$ and the last integral is convergent. As $x \to +\infty$, you have, for $b$ sufficiently great: $$ ...


1

HINT: Proportions are intuitive estimators of probabilities; i.e., to estimate $P(X \in A)$ given i.i.d. observations $X_1,...,X_n$ of $X$, consider the proportion of the $n$ observations that are in $A$: $$\hat{P}(X\in A) = \frac{1_{X_1\in A} + 1_{X_2\in A} +\ ... +\ 1_{X_n\in A}}{n}, $$ where $$1_E = \begin{cases} 1, & \text{if E occurs} \\ 0, ...


1

Here is an answer with the simplest notation I can manage. Maybe you can match ideas here with the content of the previous Answer by @r.e.s. You want to estimate the probability $p_1$ that $X = 1$ based on a sample of $n$ independent observations from the distribution of $X$. You count $Y_n$, the number of instances among $n$ in which $X = 1.$ Then $Y_n ...


1

Assume by contradiction there is a sequence $(a_n)_{n\in\mathbb{N}}$ in $A=[0,\frac{\pi}{2})$ converging to a limit $a\in B=(\frac{3\pi}{4},2\pi)$. In particular, for $$\varepsilon\stackrel{\rm def}{=} \frac{\frac{3\pi}{4}-\frac{\pi}{2}}{4}$$ there exists $N_\varepsilon$ such that for any $n\geq N_\varepsilon$, $\lvert a_n - a\rvert \leq \varepsilon$. But as ...


1

First we can observe that $n^{1/n}\to1$. In fact $n^{1/n}=e^{\frac{\ln(n)}{n}}$ and $\frac{\ln(n)}{n}\to0$. Therefore the denominator tends to $1$ while the numerator tends to $+\infty$.


1

So you are trying to prove that your function is equal to its power series representation, and as you said to do this you will use Taylors theorem and inequality. $\mathbf{Thereom:}$ If $f(x)=T_n(x)+R_n(x) , $ where $T_n$ is the $n^{th}$ degree taylor polynomial of $f$ at $a$ and $\lim_{n\to \infty} R_n(x)=0$ for $|x-a| \lt R$ , then $f$ is equal to the ...


1

Perhaps the sequence $a_1,a_3,a_5,...$ is monotone, and also $a_2,a_4,a_6,...$


1

$\textbf{HINT:}$ Try checking the secuence $\{a_{2k}\}_{k\in\mathbb{N}}$ and $\{a_{2k-1}\}_{k\in\mathbb{N}}$. One of them is crecent and the other decrecent. But it converges iff $\limsup=\liminf$


1

$a_n = \frac{n^4(x-16)^n}{4\cdot 8\cdots (4n)} $ Then $$ \bigg| \frac{a_{n+1}}{a_n} \bigg| =\frac{(n+1)^4}{n^4} |x-16| \frac{1}{4(n+1)}\rightarrow 0 $$ Hence $R=\infty$ and $(-\infty ,\infty)$.


1

For a power series, $\sum_{n \geq 1} a_n(x-x_0)^n$, then $$ R=\lim_{n\to \infty}|\frac{a_n}{a_{n+1}}| $$ In this case $$ R=\lim_{n\to \infty}\frac{n^4 4(n+1)}{(n+1)^4}=\lim_{n\to \infty}\frac{4n^4}{(n+1)^3}=\infty $$ So $R=\infty$ and $(-\infty, \infty)$ is the interval of convergence.



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