5
This is not a gaussian integral.
$$f(x) = \frac{1}{\sqrt{2\pi}} \int^{+\infty}_{-\infty} e^{-\frac{xy^4}{24}-\frac{y^2}{2}} \, dy=\frac{\sqrt{3}}{\sqrt{2\pi}}\frac{e^{\frac{3}{4 x}} }{\sqrt{x}}K_{\frac{1}{4}}\left(\frac{3}{4 x}\right)$$
Using the asymptotics of the Bessel function (since $x \to 0 \implies \frac 1 x \to \infty$), you should have
$$f(x)=1-\...
5
Indeed the radius of convergence of the function $f$ is zero, which means that the series
$$ \sum_{k=0}^{\infty} \frac{(-1)^k (4k-1)!!}{k! 24^k} z^k $$
diverges for all $z \neq 0$. Although this can be directly verified by using Stirling's approximation, a more fundamental reason for this is that $0$ is a branch point of $f(z)$. Consequently, there is no way ...
5
Since $\sin(a+b) = \sin a \cos b + \cos a \sin b$, by a Taylor expansion around $0$ we get
$$\begin{align*}
\sin\!\left(n+\frac{1}{n}\right) &= \sin n \cos \frac{1}{n} + \cos n \sin \frac{1}{n} \\
&= \sin n\left(1-\frac{1}{n^2} + o\!\left(\frac{1}{n^2}\right) \right)
+ \cos n\left(\frac{1}{n} + o\!\left(\frac{1}{n^2}\right) \right)
\end{align*}$$
...
5
First approach. By the beta integral
\begin{align*}
\frac{{\Gamma (x)}}{{\Gamma \left( {x + \tfrac{1}{2}} \right)}} & = \frac{1}{{\Gamma \left( {\frac{1}{2}} \right)}}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}B\left( {x,\tfrac{1}{2}} \right) = \frac{1}{{\sqrt \pi }}\int_0^1 {\frac{{t^{x - 1} }}{{\sqrt {1 - t} }}dt} \\ & = \frac{1}{{\...
4
You can do a Taylor development (at $\infty$) and see how that goes. Namely, write
$$
\frac{1}{n^{2/3}+(-1)^n}
= \frac{1}{n^{2/3}} - \frac{(-1)^n}{n^{4/3}} + o\!\left(\frac{1}{n^{4/3}}\right) \tag{1}
$$
using that $\frac{1}{1+x} = 1-x + o(x)$ when $x\to 0$. This means that
$$
\frac{\cos n}{n^{2/3}+(-1)^n}
= \underbrace{\frac{\cos n}{n^{2/3}}}_{a_n} - \...
4
To get an expansion for $x>>1$, let's first make substitution $y = x^{-\frac14}t$ in the integral:
\begin{align} f(x) &= \frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{t^4}{24} - \frac{t^2}{2\sqrt{x}}} \frac{dt}{\sqrt[4]{x}} = \frac{x^{-\frac14}}{\sqrt{2\pi}} \int_{-\infty}^\infty e^{-\frac{t^4}{24}} \sum_{n=0}^\infty \frac{1}{n!}\left(- \...
4
The magic words are: Stirling's formula. (check out the wikipedia article on the gamma function).
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
From $a_n<a_{n+1}<c_{n+1}<c_n$ and $b_n<c_n$ for all $n$ we can deduce
$$\begin{align*}c_{n+1}-a_{n+1} &< c_{n+1}-a_n\\
&= \frac{1}{3}a_n+\frac{1}{3}b_n+\frac{1}{3}c_n-a_n \\
&< \frac{1}{3}a_n+\frac{1}{3}c_n+\frac{1}{3}c_n-a_n \\
&= \frac{2}{3}c_n-\frac{2}{3}a_n \\
&= \frac{2}{3}(c_n-a_n)
\end{align*}$$
hence $c_n - a_n \...
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,
$$\begin{align*}
\frac{a_{n+1}}{a_n}&...
3
Dividing both the numerator and the denominator by $16^n$, you will get$$\frac{\left(\frac9{16}\right)^n+\left(-\frac12\right)^n}{\left(\frac{13}{16}\right)^n+1},$$and therefore your sequence converges to $0$.
3
Your function may be expressed in terms of the modified Bessel function of the second kind as follows:
$$
f(x)=\sqrt {\frac{3}{{2\pi x}}} e^{\frac{3}{{4x}}} K_{1/4} \left( {\frac{3}{{4x}}} \right).
$$
By the results of the paper http://dx.doi.org/10.1007/s10440-017-0099-0, for all $N\geq 0$ and $x$ with $|\arg x|<\frac{3\pi}{2}$, we have
$$
f(x) = \sum\...
3
Well, say $g(x)\geq |f_n(x)|$ for every $x$ and $n$. Then,
$$
g(x)\geq \sup_n|f_n(x)|=\begin{cases} \frac{1}{n} & |x|\in (n-1,n] \\ 0 & x=0 \end{cases}
$$
and thus,
$$
\int_{\mathbb{R}}g(x)\textrm{d}x=2\sum_{n=1}^{\infty} \frac{1}{n}=\infty
$$
and hence, $g$ cannot be integrable.
3
Take the exponential of the denominator. It can be seen that $\log\log(n) n$ is smaller than $e^n$ for large enough $n$. Since $x \mapsto e^x$ is monotonically increasing, this means the deonminator is smaller than $n$. The series can then be compared with the harmonic series.
3
Just to add a little bit to @Igor Rivin's answer - Stirling's formula gives you an asymptotic approximation for ${\Gamma(x)}$. That means that
$$
\lim_{x\to\infty}\frac{\Gamma(x)}{S(x)} = 1
$$
(where ${S(x)}$ is the Stirling formula). We write ${\Gamma(x)\sim S(x)}$ to represent this fact (of course we are assuming ${S(x)\neq 0}$). You need to use the ...
3
I think your proof is fine, but the idea behind the proof is somewhat obscured. To me, it seems the key point is to prove there is a line of positive slope that bounds $f$ from below. Indeed, if $f'(c) > 0$, then $f(x) - f(c) = f'(c_x)(x-c) \geq f'(c)(x-c)$. The right side clearly tends to $\infty$ as $x \to \infty$, hence so must $f$.
Edit, I should also ...
2
First we have
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$
for $x \in (-1,1).$
For $x=1$ the series
$$x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$
is convergent (Leibniz !) Then Abel's theorem ( https://en.wikipedia.org/wiki/Abel%27s_theorem) says that
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$
is also valid ...
2
$$
n\sum\limits_{k = 1}^{2n} {\frac{{e^{ - \frac{n}{k}} }}{{k^2 }}} = \frac{1}{2}\frac{1}{{2n}}\sum\limits_{k = 1}^{2n} {\frac{{e^{ - \frac{1}{2}\frac{{2n}}{k}} }}{{(k/2n)^2 }}} \to \frac{1}{2}\int_0^1 {\frac{{e^{ - 1/(2x)} }}{{x^2 }}dx} \mathop = \limits^{t = 1/(2x)} \int_{1/2}^{ + \infty } {e^{ - t} dt} = e^{ - 1/2} .
$$
2
Note that since for $ |a| < 1$ (Geometric series formula)
$$
\sum_{i=0}^{\infty}a^i = \frac{1}{1-a},
$$
choosing $a := \frac{1}{n+1}$, on the right hand side it follows that
$$
\frac{1}{(n+1)!}\left(\sum_{i=0}^{\infty}\left(\frac{1}{n+1}\right)^i\right) = \frac{1}{(n+1)!}\left(\frac{1}{1 - \frac{1}{n+1}}\right) = \frac{1}{n!n}
$$
2
hint
For $ (a_n)$.
we have
$$3a_{n+2}=a_{n+1}+2a_n$$
thus
$$3(a_{n+2}-a_{n+1})=-2(a_{n+1}-a_n)$$
So,
$$\alpha=\frac 23$$
For the sequence $ (c_n) $, use the canonical form
$$c_{n+1}=2-\frac{3}{3+c_n}$$
2
Well, you already gave the answer in your question.
If $\sum a_n$ converges, $\{a_n\}$ converges to zero. As we’re dealing with non negative series and $p \gt 1$, we have for all $n $ large enough
$$0\le a_n^p \le a_n$$ and therefore $\sum a_n^p$ converges.
2
The condition stated in your post isn't clear. You could have $a_n = b_n = \dfrac{1}{\sqrt{n}}\to 0$, but $\sum_{n=1}^\infty a_nb_n = \sum_{n=1}^\infty \dfrac{1}{n} = \infty$. Thus the said conditions about $a_n$ and $b_n$ should be: $a_n$ converges, $b_n$ converges, and either $\sum_{n=1}^\infty a_n$ or $\sum_{n=1}^\infty b_n$ converges. We can proceed ...
2
This is false in general, take $a_n=\frac{(-1)^n}{\log n}$ and $b_n=\frac{1}{n}$, then $\frac{b_n}{a_n}=(-1)^n\frac{\log n}{n}\rightarrow 0$, $\sum a_n$ converges but $\sum b_n$ diverges.
2
Let $c>0$ such that $\varepsilon_n A_n \geq c$ for all $n$. Then
$$\begin{align}
\mathbb{E}[(1-\varepsilon_n)^{X_n}]
&\leq \mathbb{E}[(1-\varepsilon_n)^{X_n}\mathbf{1}_{X_n \geq A_n}] + \mathbb{P}\{X_n < A_n\} \\
&\leq (1-\varepsilon_n)^{A_n}\mathbb{E}[\mathbf{1}_{X_n \geq A_n}] + \mathbb{P}\{X_n < A_n\} \\
&= 1- \left(1- (1-\...
2
$f_n\to 0$ on $\mathbb{R}$, so if the $f_n$ were dominated by an integrable function, DCT would give $$\lim_{n\to\infty}\int_{\mathbb{R}}f_n(x)\,\mathrm{d}x=0$$ whereas it's obvious that the limit is $1$.
2
For numbers of the form $-\frac{1}{n}$ the product is not defined, for the other negatives values the product diverges to $\infty$ or $-\infty$. For $x \in [0,1)$ the product diverges to $0$. For $x>1$ the product diverges to $\infty$. So the product only converges for $x=1$.
2
I'll only give a solution for $L^1(\Bbb R)$. But just by suitable normalizing, this works for other $p$ as well.
Let's first focus on $[0,1]$. Consider,
$f_{1,0}:=\chi_{[0,1]}$
$f_{2,0}:=\sqrt{2} \chi_{[0,\frac{1}{2}]}$, $f_{3,0}:=\sqrt{2} \chi_{[\frac{1}{2},1]}$
$f_{4,0}:= \sqrt{3} \chi_{[0,\frac{1}{3}]}$, $f_{5,0}:= \sqrt{3} \chi_{[\frac{1}{3},\frac{2}{3}]}...
2
Let $a_n = \frac{n^n}{n!}$. For the radius of convergence $R$ we have
$$ \frac{1}{R} = \limsup_n \sqrt[n]{a_n}. $$
Since clearly $a_n \geq 1$, also $\sqrt[n]{a_n} \geq 1$, so $\frac{1}{R} \geq 1 \implies R \leq 1$.
2
I think your instructor hoped you would notice that $n^n \ge n!$ (actually, greater when $n>1$).
At $x = 1$ the terms in this infinite sum are all at least $1$ so the series can't converge there.
2
Your assertion that $(c_k)$ is not a null sequence is correct, and that is the key step in the proof.
Noting that for $0 \leqslant i \leqslant k$ we have $\sqrt{i+1}\sqrt{k-i+1}\leqslant \sqrt{k+1}\sqrt{k-0+1}= k+1$, it follows that
$$|c_k| = \left|(-1)^k\sum_{i=0}^k \frac{1}{\sqrt{i+1}\sqrt{k-i+1}}\right|= \sum_{i=0}^k \frac{1}{\sqrt{i+1}\sqrt{k-i+1}}\...
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