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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 $\sqrt{a_n}<\sqrt{6}$.


0

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


0

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.


0

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 $1$).


0

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 ...


0

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 ...


0

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 ...


0

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 ...


0

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: $$\sum_{n\geq 1}\frac{1}{n^2}=\frac{\pi^2}{6}.$$


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.


0

$\frac{x}{n}$ 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 ...


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 ...


0

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 ...


0

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 $


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


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.


0

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 ...


3

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


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 ...


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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