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9

$$\lim_{n \to \infty} \dfrac{ 1^n +2^n+\cdots +n^n}{n^n} = \lim_{n \to \infty}\frac{n^n}{n^n}+\frac{(n-1)^n}{n^n}+\frac{(n-2)^n}{n^n}+\cdots$$ $$=\lim_{n \to \infty}1+(1-1/n)^n+(1-2/n)^n +\cdots=1+e^{-1}+e^{-2}+\cdots$$ then one can sum the geometric series.


8

Bernoulli's Inequality says that for $n\ge k$, $$ \left(1-\frac kn\right)^n $$ is an increasing sequence. Therefore, by Monotone Convergence $$ \begin{align} \sum_{k=0}^n\left(\frac kn\right)^n &=\sum_{k=0}^n\left(\frac{n-k}n\right)^n\\ &=\sum_{k=0}^n\left(1-\frac kn\right)^n\\ &\to\sum_{k=0}^\infty e^{-k}\\ &=\frac e{e-1} \end{align} $$


7

Observe that you have $$ \left|\arctan u\right|\leq|u|, \quad |u|\leq1, $$ then, switching to polar coordinates with $r=\sqrt{x^2+y^2}$, as $r\to 0$, you get $$ \left|\frac{\arctan(xy)}{\sqrt{x^2+y^2}}\right|=\frac{\left|\arctan(r^2 \sin \theta \cos \theta)\right|}{r}\leq \frac{\left|r^2 \sin \theta \cos \theta\right|}{r}\leq r. $$ The sought limit ...


4

Using the Lambert-W function you get $$ W^{-1}(\ln(x))=25\iff x=\exp(W(25)) $$ Your solution is as explicit as you can express the Lambert-W explicitly.


3

There is one more ingenious way that i want to share with you. Define functions $F(t)=t^a$ and $G(t)=t^b$ and observe their convolution, $$F \star G(t) = \int_{0}^{t} F(\lambda)G(t-\lambda) d \lambda$$ $$=> F \star G(t) = t^{a+b+1}\int_{0}^{1} y^a(1-y)^b dy, $$ (1) where we used the integral-transform determined by the relation $\lambda = ty$. In a ...


3

$-\frac{x^2+y^2}{2}\leq xy\leq \frac{x^2+y^2}{2}$ and $\arctan$ is a monotonic increasing function, so $$\frac{\arctan\frac{-r^2}{2}}{r}\leq \frac{\arctan{(xy)}}{\sqrt{x^2+y^2}}\leq \frac{\arctan\frac{r^2}{2}}{r}$$ where $r=\sqrt{x^2+y^2}$. Now, using L'Hospital's rule you get that both limits in the sandwich are $0$. Or, you can use the inequality Oliver ...


2

Use polar coordinates and L'Hospital's rule: $$\lim\limits_{(x,y) \to (0,0)} \frac{\arctan(xy)}{\sqrt{x^2+y^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{\sqrt{(r \cos(\theta))^2+(r \sin(\theta))^2}} = \lim\limits_{r \to 0} \frac{\arctan(r^2 \cos(\theta) \sin(\theta))}{r} = \lim\limits_{r \to 0} \frac{2r \cos(\theta) ...


2

HInts & useful stuff (maybe) Very interesting integral. Mathematica has no result for the DEFINITE integral. However, a close form for the INDEFINITE integral does exists and it is: $$ \int\arcsin\left(\sqrt{\frac{x}{x-a}} \right)\ \text{d}x = G(x, a) = x\arcsin\left(\sqrt{\frac{x}{x-a}}\right)+ ...


2

Indeed, the red part isn't equal to the blue one, but you should note that the sum in the blue part continues on the next line, and that $5x+20-20=5x$ to see how the first part was changed. It's almost the same for the other fraction.


2

Let's examine the first troublesome positive point, that is, $\pi$. We know that an antiderivative in the interval $(-\pi,\pi)$ is $$ f_0(x)=\frac{2}{\sqrt{3}}\arctan\frac{2\tan(x/2)+1}{\sqrt{3}}+c_0 $$ We also know that an antiderivative in the interval $(\pi,3\pi)$ is of the form $$ f_1(x)=\frac{2}{\sqrt{3}}\arctan\frac{2\tan(x/2)+1}{\sqrt{3}}+c_1 $$ Note ...


2

The standard definition of an improper integral forces you to break up the integral into separate pieces for each "bad point," and to take the limits separately, then add them together. For example, for $\int_{-1}^1\!\frac{dx}{x}$ we have \begin{align*} \int_{-1}^1\!\frac{dx}{x} &= \left[\lim_{a\to 0^-}\int_{-1}^a\!\frac{dx}{x}\right] + \left[\lim_{b\to ...


2

The mistake is marked on the screen copy below :


2

For positive $x$, $P(x)\ge 0$ and $P$ is a growing function, hence invertible, and this inverse is also growing. So for $f(m)<0,P^{-1}(f(m))$ doesn't exist, and for $f(m)\ge 0,k(m)=P^{-1}(f(m))$ varies like $f(m)$.


1

Exactness is really a property of differential forms rather than differential equations. Even though $$ M(x,y) \, dx + N(x,y) \, dy = 0 $$ represents the same differential equation as $$ f(x,y) \, M(x,y) \, dx + f(x,y) \, N(x,y) \, dy = 0 $$ (if $f(x,y) \neq 0$), it is not true that if one of the differential forms $$M \, dx + N \, dy$$ and $$(fM) \, dx + ...


1

Note that in your formula, for $\varepsilon=2-f(0)>0,$ it exists $M\in\mathbb{R}$ such that $$\forall x\in[0,+\infty),x\geq M\implies |f(x)-2|\leq 2-f(0),$$ which implies $f(0)-2\leq f(x)-2\leq2-f(0)$ and so $f(0)\leq f(x).$ Now note $m$ the minimum of $f$ on $[0,M]$ which exists according to Weierstrass theorem, and as $m\leq f(0)\leq f(x)$ for $x\geq ...


1

The comments do not seem to lead to anything fruitful. I give you the first step, and then you confirm that you have the same in your solution, OK? Write the differential equation as $$ x'-\frac{5}{y-4y^6}x=\frac{y^3}{1-4y^5} $$ Thus, an integrating factor is $$ \exp\Bigl(\int-\frac{5}{y-4y^6}\,dy\Bigr). $$ I get it to be Using this, I get the final ...


1

I do not know if the following expression is that you are looking for $$\frac12\frac{\sqrt\pi}{\Gamma(1+n)\Gamma\left(\frac32-n\right)}$$


1

We have \begin{align} {{1/2}\choose n}&=\frac{(1/2)(1/2-1)(1/2-2)\cdots(1/2-n+1)}{n!}\\ \\ &=\frac{(1/2)(-1/2)(-3/2)\cdots((3-2n)/2)}{n!}\\ \\ &=\frac{(1)(-1)(-3)\cdots(3-2n)}{2^{n}n!}\\ \\ &=(-1)^{n-1}\frac{(1)(3)(5)\cdots(2n-3)}{2^nn!}\\ \\ &=(-1)^{n-1}\frac{(2n-3)!}{2^nn!(2)(4)(6)\cdots(2n-2)}\\ \\ ...


1

As $\binom{n}{k}=\frac{n(n-1)(n-2)...(n-k+1)}{k!}$, we can write: $$\binom{\frac{1}{2}}{n}=\frac{(1/2)(-1/2)(-3/2)...(3/2 - n)}{n!}=\frac{(-1)^{n-1}(1)(3)(5)...(2n-3)}{n!2^n}$$ $$=\frac{(-1)^{n-1}(1)(2)(3)(4)(5)...(2n-3)(2n-2)}{n!2^n(2)(4)(6)...(2n-2)}=\frac{(-1)^{n-1}(2n-2)!}{2^{2n-1}n!(n-1)!}=\frac{(-1)^{n-1}(2n)!}{2^{2n}n!n!(2n-1)}$$ ...


1

$$ \begin{align} \frac{\frac12\cdot\left(-\frac12\right)\cdots\left(\frac{3-2n}2\right)}{n!} &=\frac{(-1)^{n-1}}{2^nn!}(2n-3)(2n-5)\cdots1\\ &=\frac{(-1)^{n-1}}{2^nn!}\frac{(2n-2)!}{2^{n-1}(n-1)!}\\ &=\frac{(-1)^{n-1}}{2^{2n-1}n}\binom{2n-2}{n-1} \end{align} $$


1

One useful form for that binomial coefficient is obtained by expanding it out and observing some patterns. Since $\binom{1/2}{n}=\frac{\frac{1}{2}}{n}\cdot\frac{\frac{1}{2}-1}{n-1}\cdots\frac{\frac{1}{2}-n+1}{1}=\frac{1}{2n}\cdot\frac{-1}{2(n-1)}\cdot\frac{-3}{2(n-2)}\cdots\frac{3-2n}{2}$, since there are $n-1$ factors of $-1$, and by reversing the order of ...


1

HINT: You stated the definition of the limit: For all $\epsilon >0$ there is a number $N$ so that $$-\epsilon <s_n-L<\epsilon \tag 1$$ whenever $n>N$. Now, add $L$ from both sides of $(1)$. What can you conclude?


1

You shouldn't start with what you're trying to prove. Note that the result you're trying to prove is intuitively obvious if you divide both sides by $(z-y)(z-x)$ : The average value over $[y,z]$ is larger than the average value over $[x,z]$ because the function is bigger over $[x,y]$ than $[y,z]$. Basically this is like saying the average of 2, 4, and 5 is ...


1

Write $$\frac{1}{(x+1)(x+2)^2(x+3)^3}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}+\frac{D}{(x+3)}+\frac{E}{(x+3)^2}+\frac{F}{(x+3)^3}$$ Solve for $A,B,C,D,E,F$ by comparing coefficients and integrate each term separately.


1

Hint: Write as $$\frac{1}{(x+1)(x+2)^2(x+3)^3}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^2}+\frac{D}{(x+3)}+\frac{E}{(x+3)^2}+\frac{F}{(x+3)^3}$$ The partial fraction expansion is: $$\frac{2}{x+2}-\frac{17}{8(x+3)}-\frac{1}{(x+2)^2}-\frac{5}{4(x+3)^2}-\frac{1}{2(x+3)^3}+\frac{1}{8(x+1)}$$


1

For an alternative approach write the required integral in the form $\int{\frac{1}{(x+1)(x+a)^2(x+b)^3}dx}$ then use $\frac{1}{(x+a)^2} = - \frac{d}{da} \frac{1}{x+a}$ and similarly for $\frac{1}{(x+b)^3}$ in terms of $\frac{d^2}{d b^2} \frac{1}{x+b}$, so that you have to first evaluate $I(a,b) = \int{\frac{1}{(x+1)(x+a)(x+b)}dx}$. The required integral is ...


1

By Gronwall's inequality, $f$ is identically zero.



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