9

To put that into a more reasonable (though less prone to enthusiastic upvotes) form: We have $x_1=e^{1/e}$ and $x_{n+1}=e^{x_n/e}$, and we're looking for $\lim_{n\to\infty}n\,(e-x_n)$. Letting $y_n=1-x_n/e$, some elementary algebra gives $$y_{n+1}=1-e^{-y_n}.$$ The precise value of $y_1$ is not so important, as long as it is positive. Then, $y_n$ is monotone ...


8

Presume such a function $f$ existed. Then the function $g(x)=x^2f(x)$ would be another continuous function on $[0,1]$ such that $\int_0^1 x^n g(x)dx=0$ for all $n=0,1,2\ldots$. Now, that would imply that $\int_0^1p(x)g(x)dx=0$ for every real polynomial $p$, and so, due to density of all polynomials, you would have $g(x)=0$. Thus, $f(x)=\frac{g(x)}{x^2}=0$ ...


6

The only difficulty here is justifying switching the order of integration and the summation in $$\int_0^1 \frac{dx}{1+x^3} = \int_0^1 \sum_{n=0}^\infty (-x^3)^n\, dx = \sum_{n=0}^\infty \int_0^1 (-x^3)^n\, dx = \sum_{n=0}^\infty \frac{(-1)^n}{3n+1}$$ We don't have uniform convergence of the series on the interval $[0,1]$ since the series diverges at $x=1$. ...


5

For a really simple proof, @saulspatz gave you a nice little observation. We have $$ j^2 + 1 > j^2 - 1 $$ and so $$ \frac{j^2 + 1}{j+1} > \frac{j^2-1}{j+1} = \frac{(j+1)(j-1)}{(j+1)} = j-1 $$ so $$ \frac{j^2+1}{j+1} > j-1 $$ since ${\log}$ is increasing, we have $$ \log\left(\frac{j^2+1}{j+1}\right) \geq \log\left(j-1\right) $$ can you finish it ...


5

\begin{align} \det(AB-I_3) &= (-1)^3\det(I_3 - AB)\\ &= - \det(I_2 - BA)\\ &= - (-1)^2 \det(BA-I_2) \end{align} The second equality holds due to Sylvester determinant identity. In general, if $A \in \mathbb{R}^{m \times n}$ and $B \in \mathbb{R}^{n \times m}$, then we have $$\det(AB-I_m) = (-1)^{n+m} \det(BA-I_n)= (-1)^{n+m \pmod{2}} \det(BA-I_n)$...


5

First approach. By the beta integral \begin{align*} \frac{{\Gamma (x)}}{{\Gamma \left( {x + \tfrac{1}{2}} \right)}} & = \frac{1}{{\Gamma \left( {\frac{1}{2}} \right)}}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}\int_0^1 {\frac{{t^{x - 1} }}{{\sqrt {1 - t} }}dt} \\ & = \frac{1}{{\...


4

Hints: (1) The function is continuous on any compact interval $[-a,a]$. (2) The derivative is bounded on semi-infinite intervals $(-\infty,-a]$ and $[a,\infty)$ where $a > 0$.


4

Hint: Rewrite the DE as $$\frac{dx}{dy}-\frac{x}{y}=y$$ which is easy by integrating factor method


4

We have $$ \zeta ( - 1,n + 1) = - \frac{1}{2}n^2 - \frac{1}{2}n - \frac{1}{{12}} $$ and \begin{align*} \zeta ^{(1,0)} ( - 1,n + 1) = \zeta ^{(1,0)} ( - 1,n) & + n\log n = - \frac{1}{4}n^2 + \left( {\frac{1}{2}n^2 + \frac{1}{2}n + \frac{1}{{12}}} \right)\log n + \frac{1}{{12}} \\ & - \int_0^{ + \infty } {\left( {\frac{1}{{e^t - 1}} - \frac{1}{t} ...


4

Following up on a line of reasoning suggested by MikeG and Danny Pak-Keung Chan: Let $\mu$ be the law of the $X_n$ (which is also the law of $X$). Fix $\epsilon, \eta > 0$. By Lusin's theorem applied to the measure $\mu$, there is a compact set $K \subset \mathbb{R}$ with $\mu(K) > 1-\eta$ on which which $f|_K$ is continuous. It is thus even ...


4

You can use the fact that $a^m-1=(a-1)(a^{m-1}+a^{m-2}+\cdots+1)$, with $a=\sqrt[m]{P(x)+1}$. You will get that$$\frac{\sqrt[m]{P(x)+1}-1}x=\frac{P(x)}{x\left(\sqrt[m]{P(x)+1}^{m-1}+\sqrt[m]{P(x)+1}^{m-2}+\cdots+1\right)}.$$Can you take it from here?


4

The magic words are: Stirling's formula. (check out the wikipedia article on the gamma function).


4

Let's go back to basics, where $M = [0,1]$. I'm sure you can construct a non-zero function $f: M \to \Bbb R$ such that $\int_M f(x) \ \mathrm dx = 0$.


3

As mentioned in the last comment, you need to find an $r$ such that $B(x,r)\subset [0,3)$, whenever $x\in [0,3).$ Hint: Instead of $x$, you can write $0\le x=3-\epsilon<3,$ where $0<\epsilon\le 3.$ Namely, you can express every element in $[0,3)$ as $3-\epsilon$, for some $\epsilon\in(0,3].$


3

It comes from computing the sum of a geometric series:\begin{align}1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots&=\frac1{1-1/(n+1)}\\&=\frac{n+1}n.\end{align}So,$$\frac1{(n+1)!}\left(1+\frac1{n+1}+\frac1{(n+1)^2}+\cdots\right)=\frac{n+1}{(n+1)!n}=\frac1{n!n}.$$


3

Simply False. Take $a_n=(-1)^n$ and $b_n=\frac{1}{n}$. Then $a_nb_n \to 0$ !!


3

Take $a_n=n$ and $b_n=\frac{1}{n}$. The product converges


3

Using the density of irrationals and rationals and the monotonicity of $x \mapsto \cos^2x$ on $[0,\pi/2]$, we have for any subinterval $[x_{j},x_{j+1}]$ of a partition $P$, $$\sup_{x \in [x_{j},x_{j+1}]} f(x) = \cos ^2 x_{j}, \quad \inf_{x \in [x_{j},x_{j+1}]} f(x) = 0$$ Immediately we see that the lower Darboux sum is $L(P,f) =0$ and the upper Darboux sum ...


3

Your approach is correct. The obvious choice of $\mathcal{C}$ would of course be $\mathcal{M}$. Hint: $\mathcal{M}$ must contain all unions of the form $\bigcup_{n=1}^\infty A_n$, where $A_1,A_2,\dots \in \mathcal{F}$. Is $\{\bigcup_{n=1}^\infty A_n \: | \: A_1,A_2,\dots \in \mathcal{F}\}$ a $\sigma$-algebra?


3

There are a few problems with your approach: when you write that you have to show that$$\overline{S_\sigma}<\varepsilon\ \forall\varepsilon>0,\tag1$$you don't tell was which partition $\sigma$ is, but then you acto as if it was a concrete partition. For instance, how do you know that $\frac\varepsilon2$ belongs to the partition. Besides, what you need ...


3

It should read: $\left|\dfrac{x-a}{(1+x)(1+a)}\right| < |x-a| < \epsilon$ if you let $\delta = \epsilon > 0$.


3

When you solve the recurrence $a_n=a_{n-1}+a_{n-2}$, you find that the solutions all have the form $$a_n=A\varphi^n+B\hat\varphi^n\,,$$ where $\varphi=\frac12(1+\sqrt5)$ and $\hat\varphi=\frac12(1-\sqrt5)$; the values of $A$ and $B$ are determined by the initial values $a_0$ and $a_1$. And $\varphi\hat\varphi=-1$, so, $$\begin{align*} \frac{a_{n+1}}{a_n}&...


3

$\Rightarrow)$ Suppose $A'$ is dense in $X$. Fix $x\in X$ and $\varepsilon>0$. By assumption, there is some $b\in A'$ such that $0<d(x,b)<\frac{\varepsilon}{2}$. Since $b\in A'$, there is some $a\in A$ with $0<d(a,b)<\frac{\varepsilon}{2}$. Note $d(x,a)<\varepsilon$ by the triangle inequality. This shows that $A$ is dense. The other ...


3

$$\int_{-\infty}^{\infty} |\frac {\sin (nx) \sin x}{ x^{2}}|dx$$ $$\geq \int_0^{1} |\frac {\sin (nx) \sin x}{ x^{2}}|dx$$ $$ \geq \frac 2 {\pi} \int_0^{1} |\frac {\sin (nx) }{ x}|dx $$ since $\sin x\geq \frac 2 {\pi} x$ for $0 <x<1$. Now put $y=nx $ and use the fact that $\int_0^{\infty} |\frac {\sin y} y| dy =\infty$.


3

Let us try to understand the computer-based answer of Oliver Oloa. Rational decomposition. Is known the rational decomposition of the trigonometric functions in the forms of $$\tan z = \sum\limits_{n=0}^\infty\dfrac{8z}{\pi^2\left(2n+1\right)^2 - 4z^2},\tag{R1}$$ $$\sec z = \sum\limits_{n=0}^\infty (-1)^n\dfrac{4\pi(2n+1)}{\pi^2\left(2n+1\right)^2 - 4z^2},\...


3

There is such a notion, which is used often in multivariable calculus before one introduces the Lebesgue integral. The motivation is the same - we want the volume under the graph of $f: \mathbb R^n \longrightarrow \mathbb R$. The approach of summing the area of rectangles with Riemann sums generalizes to summing the volume of higher dimensional rectangles. ...


3

This is about feedback on your proof. Your goal is to find a partition $\sigma$ such that $\overline{S}_{\sigma} (f) <\underline{S} _{\sigma} (f) +\epsilon $ but instead you are trying to use that $f$ is integrable on $[\epsilon/2,1]$ and prove that $f$ is integrable on $[0,\epsilon/2]$. This is not what you want. You want a partition $\sigma$ which ...


3

You mean $x_n = \sqrt[4]{n}$, then on $[m, 2m], m > 1, f(x) = \sqrt[4]{x}$ is differentiable thus using MVT yields: $\left|x_m - x_{2m}\right| = \left|\sqrt[4]{m} - \sqrt[4]{2m}\right| \ge |2m - m|\cdot \dfrac{1}{4\sqrt{m}} = \dfrac{\sqrt{m}}{4} > \dfrac{1}{4} = \epsilon$. Thus it’s not Cauchy ( sequence ).


3

Since proving that $$\lim_{n\to\infty} (\sin(n\alpha))^{1/n} = 1$$ may be difficult, and for some $\alpha$ it's just not true, I'd first use the comparison test. We have $$ 0 \le \left|\frac{\sin(n\alpha)}{(\ln (10))^n}\right| \le \frac{1}{(\ln (10))^n} $$ Since $$ \sum_{n=1}^\infty \frac{1}{(\ln (10))^n} $$ is convergent (by the root test), then it means ...


3

We can do a proof by “ contradiction “. Assume $s_n + t_n \to L < \infty $, then $s_n + t_n$ is bounded, say by $M$. Thus $|s_n| = |s_n + t_n - t_n| \le |s_n + t_n| + |t_n| \le M + T$. So $s_n$ is bounded, but it is unbounded since $s_n \to \infty $. Therefore $s_n + t_n \to \infty $.


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