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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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