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17

Hint (much more). One may write $$ \begin{align} \sum_{n=0}^\infty\frac{n}{(2n+1)!}&=\frac12\sum_{n=0}^\infty\frac{(2n+1)-1}{(2n+1)!} \\\\&=\frac12\sum_{n=0}^\infty\frac{1}{(2n)!}-\frac12\sum_{n=0}^\infty\frac{1}{(2n+1)!} \\\\&=\frac12\left(\frac{e+e^{-1}}2 \right)-\frac12\left(\frac{e-e^{-1}}2 \right) \\\\&=\frac12\cdot e^{-1} \end{align} $$


14

We know that $$\frac{\sinh x}{x}=\sum_{n=0}^{\infty }\frac{x^{2n}}{(2n+1)!}$$ let $x\rightarrow \sqrt{x}$ $$\frac{\sinh \sqrt{x}}{\sqrt{x}}=\sum_{n=0}^\infty \frac{x^n}{(2n+1)!}$$ $$\left(\frac{\sinh \sqrt{x}}{\sqrt{x}}\right)'=\sum_{n=1}^\infty \frac{nx^{n-1}} {(2n+1)!}$$ let $x=1$ to get what do you want


5

$$\begin{align} & \lim_{n \to\infty}\ \sum_{i=1}^{n}{(\frac{i}{n})^{2}} \\ & =\lim_{n \to\infty}\ \frac{1}{n^{2}}\sum_{i=1}^{n}i^{2}\\ & =\lim_{n \to\infty}\ \frac{1}{n^{2}}\cdot \frac{n(n+1)(2n+1)}{6}\\ & =\lim_{n \to\infty}\ \frac{1}{6}\cdot (1+\frac{1}{n})(2n+1)\\ & =\frac{1}{6}\cdot (1+ 0)\cdot \infty\\ & = \infty \end{align}$$ ...


4

Hint: a monotone decreasing sequence bounded from below converges. (Similarly a monotone increasing sequence boundes from above converges also) Of course this is a non constructive approach, as you only get the existance of a limit, and not what the limit actually is. But in some cases, this is already sufficient.


4

You can tell (intuitively) that this converges by writing down a couple of elements and trying to find a pattern: $$\frac12, \frac14, \frac18,\frac1{16}\dots$$ You should immediatelly see that the elements are becoming very small very fast. So, your intuition should tell you that the limit should be $0$. But that's only a small part of the deal. This is ...


3

There are numerous errors here. First, the general calculation should be $$\begin{align*} I_0&=\{A\subseteq D:A^c\in F_0\}\\ &=\{A\subseteq D:A^c\supseteq D_\alpha\text{ for some }\alpha\in D\}\\ &=\{A\subseteq D:A\cap D_\alpha=\varnothing\text{ for some }\alpha\in D\}\;; \end{align*}$$ $A^c\supseteq D_\alpha$ is equivalent to $A\cap D_\alpha=\...


3

Let us choose any net $t_U$ on $D=\mathcal N_{x_0}$ (ordered by reverse inclusion) such that $t_U\in U$ for each $U$. Then this net converges to $x_0$: $$t_U\to x_0$$ (In the sense of the usual definition of convergence of nets.) The modification introduced in the paper can be described as: $$ s_U= \begin{cases} t_U & U\notin C, \\ y_0 & U\...


3

Rewriting $$\frac2{n^2}\sum_{i=1}^n\sqrt{n^2-i^2}=\frac1n\sum_{i=1}^n2\sqrt{1-(i/n)^2}, $$ we recognize a Riemann sum for $$\int_0^12\sqrt{1-x^2}\,\mathrm dx.$$


3

hint: $S_n = \dfrac{2}{n}\displaystyle \sum_{i=1}^n \sqrt{1-\left(\frac{i}{n}\right)^2}$


3

Maybe you can consider $$\lim_{n\to \infty}\frac1n\sum_{i=1}^n\left(\frac {i}{n}\right)^2=\int_0^1x^2\mathrm dx=\frac13$$ Then easy to get the original limit is $\infty$.


3

No. It diverges. $ \sum_{i=1}^{n}{(\frac{i}{n})^{2}} =\frac1{n^2}\sum_{i=1}^{n} i^2 $ and $\sum_{i=1}^{n} i^2 =\frac16 n(n+1)(2n+1) \approx \frac13 n^3 $, so $ \sum_{i=1}^{n}{(\frac{i}{n})^{2}} \approx \frac13 n $.


2

It seemsĀ that you know how to handle the case where the convergence in probability is replaced by almost sure convergence. Let's do the general case. As David Mitra suggests, the key point is to extract an almost everywhere convergent subsequence. Suppose that we do not have the convergence in $\mathbb L^1$. Then there exists a positive $\delta$ and an ...


2

I presume $a_n$ are random variables. Label the events A := "$\sum_n a_n$ converges" and $B_N$ := "$\sum_n a_n > N$". We know that $\mathbb{P}[B_N] \leq C / N$ for all $N \geq 1$ for some $C \geq 0$. Note that $B_{N+1} \subset B_N$. Now, $\mathbb{P}[A^c] = \mathbb{P}[\bigcap_{N \geq M} B_N] \leq \mathbb{P}[B_M] \leq C / M$ for any $M \geq 1$. Taking $M \...


2

$\sqrt{n^2-i^2} < \sqrt{n^2} $ so $( \frac{2}{n^2} \sum_{i=1}^{n} \sqrt{n^2-i^2} ) < ( \frac{2}{n^2} \sum_{i=1}^{n} \sqrt{n^2} ) = \frac{2}{n^2} n \times n = 2 $ use comparison test. So $ \lim_{n \to \infty} \frac{2}{n^2} \sum_{i=1}^{n} \sqrt{n^2-i^2} < 2 $ so limit must be finite.


2

I think it is worth proving that $1/n\rightarrow 0$ rigorously with an epsilon-n proof at least once. After that, you can compare sequences like the one in your question to this sequence, since it is clear that $$ 2^{-n}\leq 1/n $$ Hint: use the Archimedean property.


2

If $x_n = \lceil p x_{n-1} + k \rceil$, a fixed point $x_n = a$ would have to satisfy $a-1 < p a + k \le a$, thus (if $p < 1$) $$ \dfrac{k}{1-p} \le a < \dfrac{k+1}{1-p}$$ If there were only one integer in the interval $[k/(1-p), (k+1)/(1-p))$, then that would be the only possible fixed point. But in your example, $k/(1-p) = 32000$ and $(k+1)/(1-...


1

For conditionally convergence use the Leibniz convergence theorem, the sequence is decreasing and converges to zero. But the series does not converge absolute:$$\frac{\sqrt{k}}{k+1}\geq \frac{1}{k+1}$$ which does diverge.


1

Since for $a\ne0$ zero function does not belong to $C_a$, the claim is only true for $a=0$.$\newcommand{\Ilim}{\operatorname{I-lim}}$ To prove that $C_0$ is a ring, to me the most natural way seems to be use the following facts (which are probably already known to you, if you study I-convergence): If $\Ilim f=x$ and $\Ilim g=y$, then $f+g$ is $I$-...


1

The ratio test will prove that the series is absolutely convergent if the absolute value of the result is less than $1$. This is true because the series will eventually grow smaller than an arbitrary convergent geometric series. Therefore, since we know that series a and b converges with $|x|<r_a,r_b$ respectively, we also know that as $lim_{k>\infty}$,...


1

The given sequence is LT the sequence 1/n for every n in N and so share epsilon neighborhoods about zero, no matter how small epsilon is and so converges to zero along with 1/n.


1

$VarX_n=EX_n^2=n^{2s},VarS_n=\sum_{j=1}^n j^{2s}$ so chebychev's inequality gives $P(|S_n/n|>\epsilon)\le \sum_{j=1}^nj^{2s}/(n^2\epsilon^2)\sim \int_1^nx^{2s}dx/(n^2\epsilon^2) \sim n^{2s+1}/(n^2\epsilon^2)\to 0$ iff $2s-1<0, s<1/2$. I assumed $s\neq-1/2$ for the integral approximation, but that case is analogous.


1

Let $f(x)=\frac{x^2}{\log^{\sqrt{x}}(x)}$. Then, we have $$\begin{align} \int_2^\infty f(x)\,dx&=\int_2^\infty \frac{x^2}{\log^{\sqrt{x}}(x)}\,dx \end{align}$$ Enforcing the substitution $x\to x^2$ yields $$\begin{align} \int_2^\infty \frac{x^2}{\log^{\sqrt{x}}(x)}\,dx&=2\int_\sqrt{2}^\infty \frac{x^5}{\log^x(x^2)}\,dx\\\\ &=2\int_\sqrt{2}^\...


1

For the limit, notice that $f (x)$ is bounded below by $y=0$. Next realize that for $x $ sufficiently large we have that $\ln x>x^{1/3}$ hence for large $x $ we have $$0 <f (x)<g (x)=\frac {x^2}{x^{\sqrt {x}/3}}=x^{2-\sqrt {x}/3}$$ Then notice that for $x>81$ we have that $2-\sqrt {x}/3<-1$ (it equals $-1$ at $x=81$ and is decreasing from ...


1

First step is to try the definitions. E.g. to prove $f$ is continuous: Definition: $f$ is continuous at $c$ if $\lim_{x\to c}f(x)=f(c).$ In Epsilon-Delta Language: For $\epsilon>0$, there exists $\delta>0$ such that whenever $|x-c|<\delta$, $|f(x)-f(c)|<\epsilon$. Second step (if the definitions fail) is to try some Theorems. There are many ...


1

Since $f(x)=\sqrt{1-x^2}$ is a concave function on $[0,1]$, the sequence given by $$ S_n = \frac{1}{n}\sum_{k=1}^{n}\sqrt{1-\left(\frac{k}{n}\right)^2}$$ that is trivially bounded, is also an increasing sequence by the Hermite-Hadamard inequality or by Karamata's inequality. Increasing and bounded implies convergent, hence $$ \lim_{n\to +\infty} S_n = \sup_{...


1

The Riemann sum is the most rapid tool for obtain the convergence and the closed form. But if you want another way we can consider the Abel's summation and get $$S_{n}=\sum_{k=1}^{n}\sqrt{n^{2}-k^{2}}=\int_{1}^{n}\frac{\left\lfloor t\right\rfloor t}{\sqrt{n^{2}-t^{2}}}dt $$ where $\left\lfloor t\right\rfloor $ is the floor function. Now, since $t-1\leq\...


1

To show that $f(x)$ qualifies (i.e., satisfies the hypotheses of the integral test), we need to show that $f(x)$ is continuous, positive, and decreasing. Note that the textbook says $[1,\infty)$. But it's ok for us to use $[3,\infty)$. Actually, using 1 won't even work because $f(1) < 0$. Similarly, $f(2) < 0$. If you're wondering why it's OK to ...


1

The procedure described in the post is largely right. But there is trouble at the beginning, because $\frac{1}{x(\ln x-1)}$ is not strictly decreasing. Also, the integration runs into trouble because our function is initially negative. This can all be fixed by noting that a well defined series $a_1+a_2+a_3+\cdots$ converges if and only, for example, the ...


1

For a series of positive terms of the form $$\sum_{n=1}^{\infty}\frac{1}{n^p}$$ a. it converges if $p>1$ b. it diverges if $p\le 1$ (p-series test) For eg.-$$\sum_{n=1}^{\infty}\frac{2n+3}{n^2+5}$$ diverges as for $n\rightarrow\infty$, $u_n=\frac{2n+3}{n^2+5}\approx\frac{2n}{n^2}\approx\frac{1}{n}$.


1

Here is a neat test that is relatively unknown. In any case I can think of where it would be practical to apply, I prefer a direct comparison or limit comparison, but it's certainly still interesting and useful. Consider the series $\sum_{n=0}^\infty a_n$. Suppose you have a sequence $\{b_n\}_{n=0}^\infty$ such that $\sum_{n=0}^\infty 1/b_n$ diverges. Then ...



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