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0

The correct solution is much easier than yours: $$a_{n+1}=\sqrt{6}\sqrt{a_n} > \sqrt{a_n}\sqrt{a_n} = a_n$$ since you already proved that $0 < a_n <6$, so $\sqrt{a_n}<\sqrt{6}$.

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An answer to expand on my comment. First, there is a $n_0$ such that $a_nb_n\neq0$ for $n\geq n_0$, otherwise your quotients are not well defined. Assume $n_0=1$, otherwise you start the summation from $n_0$, and you can rename the variable $k$ with $n=n_0+k-1$. Assume further that $a_1=b_1=1$, otherwise you can divide all $a_n$ by $a_1$, and all $b_n$ by ...

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 $$\sum b_N<\infty \implies \sum a_N<\infty$$

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By your hypothesis, the sequence $(a_n/b_n)$ is monotonically decreasing and bounded below (by $0$). Hence it is bounded, so $\exists M > 0$ such that $$a_n \leq M b_n$$ for $n$ large enough. Now the comparison test applies.

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Your solution is not correct. Convergent sequences do not necessarily have ratio limits less than $1$. For instance, $$\sum_{n \geq 1} \frac{1}{n^2}$$ is a reasonable, convergent sequence. But the ratio of consecutive terms approaches $1$ (and so doesn't converge to a number less than $1$).

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Let $L$ be the limit of $y_{2n+1}$, $M$ be the limit of $y_{2n}$, and let $N$ be the limit of $y_{3n}$. Then $y_{6n + 3}$ is a subsequence of both $y_{2n + 1}$ and $y_{3n}$. Thus it converges and the limit of $y_{6n + 3}$ is the same as the limit of $y_{2n + 1}$ and the limit of $y_{3n}$. Hence $L = N$. Similarly, $y_{6n}$ is a subsequence of both $y_{2n}$ ...

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 ...

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We have $y_{2n}\to L, y_{3n}\to M.$ Now $y_{6n}$ is a subsequence of both of these sequences. Therefore $L=M.$ We also know $y_{2n+1}$ converges to some $N.$ We must have $N=M$ because infinitely many of $\{3n\}$ are odd. So now all of these sequences converge to the same limit. In particular $y_{2n},y_{2n+1}$ converge to the same limit, and that impies ...

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 ...

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First we have the following asymptotic expansion \begin{align} \frac{\cos n}{\sqrt{n}+\cos n} & = \frac{\cos n}{\sqrt{n}}\frac{1}{1+\frac{\cos n}{\sqrt{n}}}\\ & =\frac{\cos n}{\sqrt{n}}\left( 1-\frac{\cos n}{\sqrt{n}}+\frac{\cos^2 n}{n}+O(\frac{1}{n^{3/2}})\right)\\ &=\frac{\cos n}{\sqrt{n}}+\frac{\cos ...

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The infinite product $$\prod \left(1-\frac{1}{k^{1+c}}\right)$$ converges absolutely, which means the series $$\sum\frac{1}{k^{1+c}}$$ converges absolutely. So a theorem says that, if you ignore any factor which is equal to zero, then the partial products $$\prod_{k=2}^n\left(1-\frac{1}{k^{1+c}}\right)$$ converge to a non-zero value. For the ...

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By considering logarithmic derivatives and the convexity of the tangent function, it is straightforward to check that: $$\forall x\in(0,1),\quad x^2\leq 2\log(\sec x)\leq x^2\tan 1$$ hence the given series is convergent by comparison with the generalized harmonic series: $$\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}.$$

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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.

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$\frac{x}{n}$ does not converge uniformly on $\Bbb{R}$. If it were uniformly convergent, then it would converge to $0$, since pointwise convergence gets you $0$. However $$\lim_{n \to + \infty} \ \left( \sup_{x \in \Bbb{R}} \ \left| \frac{x}{n} -0\right| \right)= \lim_{n \to + \infty} \ (+ \infty) = + \infty$$ so there is no uniform convergence.

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 ...

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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 ...

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We can prove this by contradiction. Suppose that the sequence does not go to zero, then there is an $\epsilon >0$ for which given any $N \in \mathbb{N}$ there is an integer $n > N$ for which $n r^n > \epsilon$. Therefore we have $$r > \epsilon^{1/n} \left( \frac1n \right)^{1/n}.$$ Since we can make $n$ as large as we like, we can then take the ...

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You can use $Abel-Pringsheim$ $Theorem$ Here Abel's (or Pringsheim's) theorem: If $\sum U_n$ is a convergent series of positive and decreasing terms, then $lim$ $nU_n = 0$. Consider the the series $\sum_{k=1}^{n} r^n$ where $0 \leq r<1$ apply $Abel-Pringsheim$ $Theorem$

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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: $$0\leq n\cdot e^{-sn} = \frac{n}{\left(e^{\frac{s}{2}n}\right)^2}\leq\frac{n}{\left(1+\frac{s}{2}n\right)^2}\leq\frac{4}{s^2 n}.\tag{2}$$

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.

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We can use Cauchy-Schwarz inequality. In fact $$\left(\sum_{n\geq1}\frac{a_{n}^{1/2}}{n}\right)^{2}=\left(\lim_{N\rightarrow\infty}\sum_{n\leq N}\frac{a_{n}^{1/2}}{n}\right)^{2}=\lim_{N\rightarrow\infty}\left(\sum_{n\leq N}\frac{a_{n}^{1/2}}{n}\right)^{2}\leq\lim_{N\rightarrow\infty}\sum_{n\leq N}a_{n}\lim_{N\rightarrow\infty}\sum_{n\leq ... 0 Let \Omega_X be the set of full measure on which X_n \rightarrow X. Define \Omega_Y likewise. Note that \Omega_0:=\Omega_X \cap \Omega_Y has full measure as well. On \Omega_0, X_n \rightarrow X AND X_n \rightarrow Y. Hence X=Y a.s due to uniqueness of limit. 1 Suppose that X_n\to X almost surely as n\to\infty and X_n\to Y almost surely as n\to\infty. Then there exists \Omega'\subset\Omega such that \Pr(\Omega')=1 and for each \omega\in\Omega'$$ |X_n(\omega)-X(\omega)|\to0 $$as n\to\infty. Similarly, there exists \Omega''\subset\Omega such that \Pr(\Omega'')=1 and for each \omega\in\Omega'' ... 1 For the first part, you have established that -5 and -3 are not part of the radius of convergence, therefore the interval is (-5,\,3), and not [-5,\,3] (which would include -5 and -3). Same logic for the second part. You've shown that the series converges for -5 < x < -3 but diverges for -5 and -3, therefore, it is the open interval ... 0 Define x_n=\dfrac{10^n-1}{3.10^n}. Then establish \left|x_n-\dfrac13\right|=\left|\dfrac{10^n-1}{3.10^n}-\dfrac13\right|. And so, \left|\dfrac{10^n-1}{3.10^n}-\dfrac13\right|=\left|\dfrac{10^n-1-10^n}{3.10^n}\right|=\left|\dfrac{-1}{3.10^n}\right|=\left|\dfrac{1}{3.10^n}\right|\lt\epsilon\;\forall n\ge N(\epsilon) 0 You did no justify your value for x_n. The value of a decimal (base 10) string with infinite number of digits 3 behind the period is:$$ (0.333\cdots)_{10} = \sum_{k=1}^{\infty} 3\cdot 10^{-k} = 3 \sum_{k=1}^{\infty} 10^{-k} = 3 \sum_{k=1}^{\infty} \left(\frac{1}{10}\right)^k = 3 \lim_{n\to \infty} S_n $$for the partial sums$$ S_n = \sum_{k=1}^{n} ...

0

The $N$ comes in as an "interval around infinity." The definition says that there is an $N$ such that for all $n> N$ (i.e., for all $n\in(N,\infty)$), then the final inequality holds. What you have done so far is say that it must be the case that $\frac{1}{3}10^{-n}<\varepsilon$, which is the same as saying $n>-\log_{10}(3\varepsilon)$. If you ...

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

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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 $Y_n ... 0 Hints and remarks. No. Let$a_n=b_n=1/n^2$. Both series are absolutely convergent, while$\sum_{n=1}^{\infty}1$is obviously divergent. What do we know about the radius of convergence of$(\sum a_nx^{n})'$, knowing the radius of$\sum a_nx^{n}$?$a_n$'s are close to zero for large$n$and $$\lim_{x\to0}\frac{\tan x}{x}=1.$$ 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 ... 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 $$1_E = \begin{cases} 1, & \text{if E occurs} \\ 0, ... 0 Note that$$\left|a-a_n\right|<\varepsilon$$Implies:$$\left|a-a_n\right| \leq \varepsilon$$On the other hand:$$\left|a-a_n\right| \leq \varepsilon$$Implies:$$\left|a-a_n\right| < 2\varepsilon$$0 In your sums, call the terms in brackets c(k,n). Let's think of c(k,n) as defined for all k,n\in \mathbb {N}, with c(k,n) = 0 for k>n. So we are dealing with \sum_{k=1}^{\infty}c(k,n)q^k. Now |c(k,n)|\le 2 for all k,n. And for fixed k,\lim_{n\to \infty}c(k,n) = 0. Because \sum_{k=1}^{\infty}q^k < \infty, the dominated convergence ... 1 This is a famous sum & there are multiple ways of approaching this. Here is one way via a geometric sum: We have the well known sum \begin{equation*} 1+x+x^2+\cdots =\frac{1}{1-x}. \end{equation*} Setting x=e^{i\theta},~0<\theta<2\pi~(x\neq 1) gives \begin{equation*} 1+e^{i\theta}+e^{2i\theta}+\cdots ... 1 Assume by contradiction there is a sequence (a_n)_{n\in\mathbb{N}} in A=[0,\frac{\pi}{2}) converging to a limit a\in B=(\frac{3\pi}{4},2\pi). In particular, for$$\varepsilon\stackrel{\rm def}{=} \frac{\frac{3\pi}{4}-\frac{\pi}{2}}{4}$$there exists N_\varepsilon such that for any n\geq N_\varepsilon, \lvert a_n - a\rvert \leq \varepsilon. But as ... 2 Thomas Andrews is right when he says that there are no general rules. Often this type of problem, we must use a more intuitive method. I know the p-series for p = 2. The resolution method in this series comes from Euler himself. (1) \sin(x) = x - x^3/3! + x^5/5! -x^7/7! + ... (Taylor series) (2) \sin(x)/x = 1 - x^2/3! + x^4/5! -x^6/7! + ... (divided ... 3 No, there is no general formula yet to know the sum of$$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!}}$$where B_n is a Bernoulli number. However for odd integers we still don't have a general formula, if you are interested in ... 1 Hint. The integrand is a continuous function on (0,\infty), thus the potential problems are near 0 and near \infty. As x \to 0^+, you have, for b sufficiently near 0^+:$$ \int_0^b \frac{1}{e^{\sqrt{x}}-1}dx \sim \int_0^b \frac{1}{\sqrt{x}}dx $$and the last integral is convergent. As x \to +\infty, you have, for b sufficiently great:$$ ... 0 Follow the hint given in the question: use the argument in Sec. 5.11, use the Banach-Steinhaus Theorem in exactly the same way with the change of the linear functionals to${\Lambda _n}f = \frac{1}{{{\lambda _n}}}{s_n}(f;0)$. The proof is almost word for word there. Again the hint suggest a better estimate for the 1-norm of the Dirichlet kernel Dn. In the ... 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 $$\text{var}(x_i) - 2 \text{cov}(x_i, x_j) + \text{var}(x_j) = ... 0 At least we can build an example where the last convergence does not hold. Let X=X^* be a separable (for simplicity) Hilbert space, take x_n=e_n the canonical basis. Y=\Bbb R. Take also T_n=e^*_n - the canonical basis in X^* (by Riesz representation, we can say that T_n=x_n and T_ny = (e_n,y)_X). Then \forall x\, T_nx\to0 (so T=0), ... 0 You can use continuity from above: if \nu is a finite measure on (X,\mathcal A) and A_n \searrow \emptyset, then \lim_{n \to \infty} \nu(A_n) = 0. In this case, since f \in L^1(\mu), the set function$$ \nu(A) = \int_A f^+ \, d\mu$$is a finite measure so you have$$\lim_{n \to \infty} \int_{A_n} f^+ \, d\mu = 0.$$Likewise you have$$ \lim_{n \to ... 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. 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$1 Perhaps the sequence$a_1,a_3,a_5,...$is monotone, and also$a_2,a_4,a_6,...$1 First we can observe that$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 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. 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 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.\$

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