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1

Let's try this: $(n+1)^{a}(m+1)^{b}=2\cdot n^{a}m^{b}$ Obviously either $(n+1)$ or $(m+1)$ must be even but not both. Take $(n+1)=2^{j}\cdot r;(n+1)^{a}=2^{a\cdot j}r^{a}$ Now $n+1$ is relatively prime to $n$ so if $a\cdot j>1$ then $m=2^{k}s$ and we have $a\cdot j-b\cdot k=1$ Thus $(a,b),(a,j),(j,k),(j,b)$ are all relatively prime. When we ...


7

Let $P(B)$ be the probability that Bob wins the game. We will have three cases based on the first roll of the game: X) First roll sums to $12$ Y) First roll sums to $7$ Z) First roll sums to anything other than $12$ or $7$. As these cases are disjoint, we can write $P(B)$ as a sum of conditional probabilities, namely $$ ...


10

Let's call the probability that Bob wins be $x$. On the first roll of the two dice, there are three options: The dice show a sum of $12$ (probability $\frac{1}{36}$) and Alice would win immediately. The dice show a sum of $7$ (probability $\frac{1}{6}$) and this is a little more complicated (see next tree diagram, where let the probability that Bob would ...


1

Yes! If $x_{nk}$ doesn't tend to $\infty$, we can't have $x_n \to \infty$. ( $x_n \to l$ iff. $\forall (x_{k_n})_{n \ge 1}$ a subsequence of $(x_n)_{n \ge 1}$ tends to $l$)


2

As the other answers have pointed out (at least within the framework of ZFC), the answer is "proper class many." Let me just point out that this isn't the end of the story. We can ask whether two proper classes have the same size, by asking about the existence of a definable bijection, etc. In this sense the class of ordinals - or equivalently of ...


-1

The different sizes of infinities are called transfinities. They can be captured as mathematical sets via formalized definitions. The cardinality (number of) transfinities is not itself a transfinity, it is infinity. It cannot be captured as a set. It cannot be formalized by a mathematical definition. In that sense it is truly infinity which just means an ...


1

We don't know if space is infinitely divisible, but if it is, then it has uncountably many points, because if $(x_n)$ is any infinite sequence of points, and if a sequence of regions of space $(S_n)$ is constructed recursively so that $S_{n+1} \subset S_n \setminus \{x_n\}$ ($n = 0, 1, 2, \ldots$), then this nested sequence contains at least one point that ...


0

I guess in a sense it depends if you believe the universe is inherently digital or analog. If analog, there is infinite detail, so the surface area of any actual physical object would be infinite.


-3

Set of stars is infinite in reality. Infact it is countably infinite. Another example of infinite set is "set of all points in a line segment is an infinite set" See this it says the that set of integers is countably infinite whereas set of real numbers is uncountably infinite. In countably infinite sets you cannot count the total number of elements but ...


1

If you accept that one can form infinite sequences from say the set of symbols $\{a,b\}$, then I would say that the set of all such sequences gives a fairly nice example of an uncountable set; as shown by Cantor's diagonal argument. This may be simpler than the uncountability of the reals, or of the interval $[0,1]$ as you don't have to worry about some ...


8

That depends. If by "realistic" you mean something that has to do with physical reality, then I defy you to come up with a set which has exactly $200^{200^{200}}$ elements. If by "realistic" you mean something which comes up naturally in mathematics, then $\Bbb R$ is an uncountable set. As for explaining the difference between them? That's not very easy, ...


1

The idea consists of the following: consider the unit sphere $\mathbb{S}^2 \subset \Bbb R^3$, and look at $$\Bbb C \equiv\Bbb R^2 \cong \{(x,y,0)\mid x,y \in \Bbb R \}\subset \Bbb R^3$$ Consider the north pole $N= (0,0,1)$. Consider the stereographic projection from $\Bbb S^2 \setminus N$ to $\Bbb C$. Then you take a object $\infty \not\in \Bbb C$ and ...


9

Whether or not things are "undefined" largely depeneds on what framework you are working in. If we are working in the naturals, we might say that $3-5$ is undefined. There are many systems where it makes sense to assign $\frac{n}{0}$ some value. In this particular example, it is defined to be complex infinity, which can be thought of as follows: suppose we ...


3

Sometimes it is useful in complex analysis to consider the complex numbers plus the "point at infinity". See this wiki article for details: Riemann Sphere


1

Your last step is incorrect. In fact, $$ \frac{(n+1)^{n+1}}{n^n(n+1)} = \frac{(n+1)^n}{n^n} = \left(\frac{n+1}{n}\right)^n = \left(1+\frac1n\right)^n $$ which you might recognize. In your last step, on the bottom you replaced $n+1$ with $n$, which is okay here because their ratio is $\frac{n+1}{n} = 1+\frac1n \to 1$; but on the top you replaced ...


1

$$\frac{(n+1)^{n+1}}{(n+1)!}\frac{n!}{n^n}=\frac{(n+1)(n+1)^n}{(n+1)n!}\frac{n!}{n^n}=\frac{(n+1)^n}{n^n}=(1+\frac{1}{n})^n\rightarrow e>1$$


2

Hint: Show by induction or inspection that $n^n \geq n!$. Then your terms do not converge to zero so the series necessarily diverges.


1

The sphere only contacts the floor at one point, so there's zero surface area. Here's an argument that hopefully blends the physical intuition and mathematical rigor you want: For the sake of contradiction, imagine what would happen if there were two or more points of intersection: Viewed in the floor, these two points are connected by a straight line ...


10

$$\lim_{n\to\infty} \frac{3^n}{2^n+3^n}=\lim_{n\to \infty}\frac{1}{(\frac{2}{3})^n+1} =1.$$Since $\frac{2}{3}<1$ , so $(\frac{2}{3})^n\to 0$ as $n\to \infty$.


2

Use Stirling's Approximation: $$n! \approx \sqrt{2 \pi n} .n^n e^{-n}$$ $$ \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} $$ $$ =\lim_{n\to\infty} \frac{2 \pi n .n^{2n} e^{-2n}}{\sqrt{2 \pi 2n} .(2n)^{2n} e^{-2n}} $$ $$ =\lim_{n\to\infty} \frac{\sqrt{\pi n}}{4^n}=0$$


6

As all terms are positive, we have $$0 \leq \frac{(n!)^2}{(2n)!} = \frac{n!}{2n \cdot \dots \cdot (n+1)} = \prod_{k=1}^n \frac{k}{k+n} \leq \prod_{k=1}^n \frac{1}{2} = \left(\frac{1}{2}\right)^n$$ So then as $$\lim\limits_{n\rightarrow\infty} \left(\frac{1}{2}\right)^n = 0$$ It follows that $$\lim\limits_{n\rightarrow\infty} \frac{(n!)^2}{(2n)!} = 0$$ ...


8

Hint: for combinatorial reasons (in how many ways can you choose $n$ objects from $n$ pairs of objects?), $$\binom{2n}n\ge2^n\ ,$$ so $$\frac{(n!)^2}{(2n)!}=\frac1{\binom{2n}n}\le\frac1{2^n}\ .$$


2

Since each term is bounded above by $2^{-n}$, I guess the answer must be zero.


4

Recalling the Stirling's approximation $$n!\sim\sqrt{2\pi}n^{n+1/2}e^{-n} $$ we have $$\sqrt{2\pi}\lim_{n\rightarrow\infty}\frac{n^{2n+1}e^{-2n}}{\left(2n\right)^{2n+1/2}e^{-2n}}=0. $$


0

In $(-\infty,0)$ the function is identically the constant $-1$. The limit of a constant is this constant.


2

Another way to do: Let $a_n=\frac{10^n}{n!}$. Clearly $$a_{n+1}=\frac{10}{n+1}a_n$$ and hence $\{a_n\}$ is positive and decreasing when $n\ge 10$. By the bounded Monotone Principle, $\lim_{n\to\infty}a_n=L$ exists. Thus $$ L=\lim_{n\to\infty}a_{n+1}=\lim_{n\to\infty}\frac{10}{n+1}a_n=0\cdot L=0.$$


2

If $n>20$, then $$ 0<a_n=\frac{10^n}{n!}<\frac{10^{20}}{20!} \left(\frac{10}{20}\right)^{n-20}=\frac{20^{20}}{20!}\cdot\frac{1}{2^{n}}\to 0, $$ as $w^n\to$, whenever $\lvert w\rvert<1$.


7

$$\frac{10^n}{n!} = \frac{10\cdot 10 \cdot 10 \cdot 10 \cdots 10}{1 \cdot 2 \cdot3 \cdots n} =\frac{10^{11}}{11!}\frac{10}{12}\frac{10}{13}\cdots\frac{10}n < \frac{10^{11}}{11!}\left(\frac{5}{6}\right)^{n-11} \to 0$$ as $n \to \infty.$


0

@kmitov gives a very good prove by using ratio test. () Also, you could always remember that factorial grows faster then exponential function. So the limit goes to 0. For a prove: Set $a_n=10^n/n!$ and take arbitrary $\epsilon>0$, you can always find a $N$ large enough such that for all $n>N$ such that $$ a_n<\epsilon $$ Together with the fact ...


4

The series $\sum_{n=0}^\infty \frac{10^n}{n!}$ is convergent and the sum is $e^{10}$. The convergence can be checked by ratio test $$\lim_{n \to \infty} \frac{10^{n+1}}{(n+1)!}\frac{n!}{10^n}=\lim_{n \to \infty}\frac{10}{n+1}=0<1$$ By the necerssery condition for the convergence of an infinite series the common term must go to 0.


0

suppose lim$a_n$=x, Then lim$a_n$+$_1$=x also now $x=√(2+x)$ squaring it you will get $x^2$=$x+2$ solving this quadratic equation we get $x$=-1,2 $x$=-1 is absurd as sequence is increasing and starts from $√2$ we have only choice $x=2$, which is the limit of the sequence


0

Once you have proved that the sequence is convergent, then, making $n \to \infty$ in the recurrence relation defining $a_n$, the limit $\ell$ satisfies $$ \ell =\sqrt{2+\ell}, \quad \ell>0. $$ equivalently $$ \ell^2 =2+\ell, \quad \ell>0. $$ which may be solved to obtain $\ell$.


0

hint: $0\leq |a_nb_n| \leq K|a_n|, n \geq N_0$.


3

"Two completely random numbers" is not a well-defined probability distribution, even when restricted to positive real numbers. The actual probability distribution must be something more specific, for example, if all positive real numbers are possible choices then the exponential distribution is a possibility. But notice that with an exponential ...


1

You are right and wrong. First off you can't give a uniform probability to a countably infinite range (the discrete case), but you can for an uncountable domain (the continuous case). But in the countable case you can always give it some probability distribution and the probability that the first one is greater than the second is never going to be exactly ...


1

In order to make a guess, you should begin by calculating some partial sums: $$\begin{align*} \sum_{i=1}^1\frac{i}{(i+1)!}&=\frac12\\\\ \sum_{i=1}^2\frac{i}{(i+1)!}&=\frac12+\frac2{3!}=\frac12+\frac13=\frac56\\\\ \sum_{i=1}^3\frac{i}{(i+1)!}&=\frac56+\frac3{4!}=\frac56+\frac18=\frac{23}{24}\\\\ ...


2

We have:$$\dfrac{i}{(i+1)!} = \dfrac{(i+1)-1}{(i+1)!} = \dfrac{i+1}{(i+1)!} - \dfrac{1}{(i+1)!} = \dfrac{1}{i!}-\dfrac{1}{(i+1)!} \Rightarrow S_n = 1-\dfrac{1}{(n+1)!}\Rightarrow S=\displaystyle \lim_{n\to \infty} S_n = 1$$


1

You can use $e^{-1/x} = 1 - \tfrac 1x + O(\tfrac 1{x^2})$: $$\begin{align} \lim_{x\to\infty} (x-1)e^{-1/x}-x & = \lim_{x\to\infty} (x-1)(1 - \tfrac 1x + O(\tfrac 1{x^2}))-x \\ &= \lim_{x\to\infty} (x-1) + -\tfrac 1x(x-1) + O(\tfrac 1{x^2})-x \\ &= \lim_{x\to\infty} x-1 + -1 +\tfrac 1x + O(\tfrac 1{x^2})-x \\ &= \lim_{x\to\infty} -2 +\tfrac ...


4

$$(x-1)e^{-1/x} - x = x(e^{-1/x} - 1) - e^{-1/x}$$ $$\lim x(e^{-1/x} - 1) = \lim \frac{e^{-1/x} - 1}{1/x}$$ Use L'hopital's rule.


2

You can extend the real numbers like that, as an ordered set, but you cannot extend the basic algebraic operations to that extended set in a way that allows you to do things like cancel terms from both sides of an equation. You need to be able to do those types of algebraic operations to prove the things that result in absurdities. To do those kinds of ...


2

You're talking specifically about the projective compactification of the number line, and the issue is that in that setting it doesn't make sense to multiply numbers together. Or, more specifically, you just cannot multiply $0 = [0:1]$ by $\infty = [1:0]$, since this would result in $[0:0]$ which does not represent a point on $\mathbb{P}^1$ at all. Read ...


1

I think that Taylor series could be a simple solution. Considering that, for small $x$, $$\tan(x)=x+\frac{x^3}{3}+O\left(x^4\right)$$ $$n \tan\big(\frac{\pi}n\big)=n \Big(\frac{\pi }{n}+\frac{\pi ^3}{3 n^3}+\cdots)=\pi+\frac{\pi ^3}{3 n^2}+\cdots$$ For the second (without using the good hint user222031 provided in comments) ...


0

My understanding of the Hilbert Hotel is that the hotel has $n$ rooms. Since the number of rooms is infinite a new guest will get the room $n_{+1}$. So if $k$ people arrive the will stay in the rooms $n_{+1},...,n_{+k}$, and there is no need for the other guests to move.


4

What you are trying to prove is that "if $x$ is large enough, then $\frac{1}{x}$ gets close enough to $0$." Approaching infinity represents the concept of allowing $x$ to be arbitrarily large. Here is how to state the proof, with some comments along the way. Let $\varepsilon>0$ be some arbitrary real number (this will represent "error" or "distance ...


2

Let $f:\Bbb R\setminus\{0\}\to\Bbb R$ be $f(x)=1/x$. You have conjectured that $$ \lim_{x\to\infty}f(x)=0 $$ Rigorously, this means that for every $\varepsilon>0$ there exists an $M$ such that $$ \left\lvert\frac{1}{x}\right\rvert<\varepsilon $$ whenever $x>M$. Can you prove this?


1

First, as $x \to \infty$, $\frac1{x} \to 0$. Second, for $|x| < \pi/2$, $|\sin(x)| \le |x|$. Therefore, for $x > \frac{2}{\pi}$, $|\sin(\frac1{x})| \le \frac1{x} $, so $\sin(\frac1{x}) \to 0$.


1

If I make $x$ bigger and bigger, what happens to $\frac 1 x$? I assume it's clear to you that it goes towards zero. Now if I make the $x$ in $\sin x$ go towards zero, what happens to $\sin x$? What I'm getting at is that $$\lim_{x\to \infty} \sin\left(\frac 1 x\right)=\sin\left(\lim_{x\to\infty}\frac 1 x\right)$$ Which should be obvious if you have a ...


1

The standard proof that the real numbers are uncountable does not need to be phrased as a proof by contradiction. To say that $\mathbb{R}$ is uncountable is to say that if $f : \mathbb{N} \to \mathbb{R}$ is a map, then it is not a bijection. The standard proof in fact proves that $f$ is not a surjection by explicitly exhibiting a real number which is not in ...


0

Consider $$\oint_C dz \frac{z \, e^{i 4 z}}{(z-2)^2+4} $$ where $C$ is a semicircle of radius $R$ in the upper half plane, with diameter along the real axis. The contour integral is then equal to $$\int_{-R}^R dx \frac{x \, e^{i 4 x}}{(x-2)^2+4} + i R \int_0^{\pi} d\theta \, e^{i \theta} \frac{R e^{i \theta} \, e^{i 4 R e^{i \theta}}}{\left (R e^{i ...


0

We want the $$\text{Imag}\left(\int_{-\infty}^{\infty} \dfrac{x\exp(4ix)dx}{x^2-4x+8}\right)$$ Consider the contour integral $$f(R) = \int_{[-R,R]\cup \Gamma_R} \dfrac{z\exp(4iz)dz}{z^2-4z+8} = \int_{-R}^R \dfrac{z\exp(4iz)dz}{z^2-4z+8} + \int_{\Gamma_R} \dfrac{z\exp(4iz)dz}{z^2-4z+8}$$ where $\Gamma_R$ is a semi-circle of radius $R$ on the upper half plane. ...



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