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Hint: $$\dfrac{d}{dt} \sum_{n=0}^\infty t^n = \sum_{n=1}^\infty n t^{n-1}$$ for $|t|<1$.

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This is not an answer but it is too long for a comment. If we consider the more general term $$a_n = \frac{(n!)^2 k^{n}}{(2n)!}$$ and using Stirling approximation, just as Clement C. did in his answer, we have $$a_n \operatorname*{\sim}_{n\to\infty} \left(\frac k 4 \right)^n \sqrt{\pi n}$$ and $a_n$ goes through a maximum value when $n=\frac 1{2 \log\left(\... 0$(1)$says that for each positive$\delta$there is an$x$with a certain property;$(2)$just makes that claim only for those values of$\delta$that can be written in the form$\frac1n$for some positive integer$n$. If the claim holds for all positive$\delta$, then it certainly holds for all$\frac1n$. In more detail,$(1)$says that there is some$\...

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Notice that for any $n \in \mathbb N$, we have that: \begin{align*} (2n)! &= \color{red}{(2n)}(2n - 1)\color{red}{(2n - 2)}(2n - 3)\color{red}{(2n - 4)}(2n - 5) \cdots \color{red}{(4)}(3)\color{red}{(2)}(1) \\ &\leq \color{red}{(2n)}(2n)\color{red}{(2n - 2)}(2n - 2)\color{red}{(2n - 4)}(2n - 4) \cdots \color{red}{(4)}(4)\color{red}{(2)}(2) \\ &= ...

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Method 1. As a first remark, Stirling's approximation will do the job: $$a_n = \frac{n!^2 4^{n}}{(2n)!} = \frac{4^n}{\binom{2n}{n}} \operatorname*{\sim}_{n\to\infty} \sqrt{\pi n}$$ and the series trivially diverges (the general term does not even converge to zero). Method 2. A first simpler idea (than Stirling's): recognize a Binomial coefficient. For ...

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The reason they mention that $x\in(-1,1)$, which is I think the main source of your confusion, is so that you can perform things like integration and differentiation as you would a polynomial, termwise. As long as $x$ is inside the radius of convergence, $$P'(x)=1-x^2+x^4-x^6+\cdots$$ Which as others have mentioned converges to $$\frac{1}{1-(-x^2))}=\frac{... 0 We can approximate in the following manner:$$F(x)\approx\sum_{n=0}^\infty e^{-x(n+1/2)}=\frac{e^{-x/2}}{1-e^{-x}}$$1 For each particular b, it will have a closed form: since the \sin term depends only on n \mod (2b), this reduces to the sum of 2b geometric series. I doubt that there is a closed form expression as a function of a and b. 1 P'(x)=1-x^2+x^4-x^6+\ldots is a geometric series, and converges to \frac{1}{1+x^2} for |x|<1. 0 Regarding the main question: I suppose "not exactly". However, since every cube can be solved in 20 moves or less, one would "only" have to do the computations up to m=20. For m=20 the question simply degenerates to What are the possible orders of elements of the Rubik group? How many elements of a given order exist? I am sure these two questions ... 2 Taking the absolute value of your algebra gives x^2. Since the ratio test looks for when this is <1, we ask ourselves when is |x^2|<1. The answer is when -1<x<1, and if |x|>1 it does not converge. This suffices to find the radius; it is 1. If you want the interval of convergence as well, then you need to test when |x|=1. ... 2 Suppose \{a_n\} is the sequence of terms of this series. Then it is clear that \lim \sup (|a_n|)^{1/n} = 1  so that the radius of convergence is 1. 1 You do not really need the integral test to prove it. Now that Landau's notation is your forte, you may simpy notice that$$ \text{arctanh}\frac{1}{k}=\frac{1}{2}\log\left(\frac{1+\frac{1}{k}}{1-\frac{1}{k}}\right)=\sum_{n\geq 0}\frac{1}{(2n+1)\,k^{2n+1}}=\frac{1}{k}+O\left(\frac{1}{k^3}\right)\tag{1} $$implies:$$ H_n = \sum_{k=1}^{n}\frac{1}{k} = O(1)+\...

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Whether the sum converges is irrelevant with the first infinite terms. So we can focus on the terms when n tends to infinity. In this sense, $$\frac{2 \sqrt{n}+1}{n^2+n+1} \sim n^{-3/2}$$

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You have already found that: $\forall \ n, a_n+b_n+c_n=1$ Without loss of generality, suppose that: $a_{n}\leq b_{n}\leq c_{n}$. Then: $a_{n+1}=a_{n}^{2}+2b_{n}c_{n}=a_{n}^{2}+b_{n}c_{n}+b_{n}c_{n}\leq a_{n}c_{n}+b_{n}c_{n}+c_{n}c_{n}=c_{n}(a_{n}+b_{n}+c_{n})=c_{n}$. Similary, $b_{n+1}=b_{n}^{2}+2a_{n}c_{n}\leq b_{n}c_{n}+a_{n}c_{n}+c_{n}^{2}=c_{n}(a_{n}+... 0 You can use Cauchy-Schwarz inequality with$u_i = \sqrt{a_i}$and$v_i=\frac{1}{\sqrt{a_i}}$. 4 You're on the right track, but you gave up too much in the numerator. Instead, observe that $$\frac{2\sqrt{n}+1}{n^2+n+1}\leq \frac{2\sqrt{n}+1}{n^2}\leq \frac{3\sqrt{n}}{n^2}=\frac{3}{n^{\frac{3}{2}}}$$ and$\sum_{n=1}^{\infty}\frac{3}{n^{\frac{3}{2}}}$converges. 1 The function $$f(x)=\frac{1}{x}$$ Is convex on$(0,\infty)$, so for$a_1,\ldots,a_n>0$we have $$\frac{n}{a_1+\cdots+a_n}\leq\frac{1}{na_1}+\cdots+\frac{1}{na_n}.$$ Rearranging, we obtain the result. 0 We let$a_{i+n}=a_i$for$i\in\{1,2\dots n\}$.What you have is$\sum_\limits{i=1}^n (\frac{a_1}{a_{1+i}}+\frac{a_2}{a_{2+i}}+\dots \frac{a_n}{a_{n+i}}).$By the rearrangement inequality each of these summands is greater than$1+1+\dots +1=n0 This is not a complete answer but rather a hint for a attempt. Note that $$\forall n,\;a_{n+1}+b_{n+1} = 1-c_{n+1}$$ so \begin{eqnarray*} \forall n,\;a_{n+1} &= \frac{1}{2}[(a_{n+1}-b_{n+1}) + (a_{n+1}+b_{n+1})]\\ &= \frac{1}{2}\left[(a_1-b_1)\prod_{i=1}^n(1-3c_i) + 1-c_{n+1}\right] \end{eqnarray*} and similarly \begin{eqnarray*} \forall n,\;b_{n+1}... 1 We can use the standard formula with a slight variation: We obtain \begin{align*} \left(\sum_{k=0}^\infty a_kx^{2k}\right)\left(\sum_{l=0}^\infty b_lx^l\right) &=\sum_{n=0}^\infty\left(\sum_{{2k+l=n}\atop{k,l\geq 0}}a_kb_l\right)x^n\tag{1}\\ &=\sum_{n=0}^\infty\left(\sum_{k=0}^{\left\lfloor\frac{n}{2}\right\rfloor}a_kb_{n-2k}\right)x^n\tag{2} \... 1 For a continued fraction of the following form to converge: $$\cfrac { 1 }{ a_1+\cfrac { 1 }{ a_2+\cfrac { 1 }{ a_3+\cfrac { 1 }{ a_4+\dots } } } }$$ $$a_k >0$$ The following sum has to diverge: $$a_1+a_2+a_3+a_4+\cdots$$ Thus, even the following continued fraction converges: $$\cfrac { 1 }{ 1+\cfrac { 1 }{1/2+\cfrac { 1 }{ 1/3+\cfrac { 1 }{1/4+\... 4 The modified Bessel functions of the first kind fulfill the recurrence relation:$$ I_n(\alpha)=\frac{\alpha}{2n}\left(I_{n-1}(\alpha)-I_{n+1}(\alpha)\right)\tag{1}$$due to the integral representation:$$ I_n(\alpha) = \frac{1}{\pi}\int_{0}^{\pi}\cos(nx)\,e^{\alpha\cos x}\,dx \tag{2}$$and the cosine addition formulas. A straightforward consequence of (1) ... 1 This particular one is I_1(2)/I_0(2) where I_0 and I_1 are modified Bessel functions of the first kind, of orders 0 and 1. 0 In addition to @ncmathsadist, hoping, this isn't homework \sum _{i=1}^n \left(\frac{\left(2-\sqrt{5}\right)^i}{2 \sqrt{5}}-\frac{\left(\sqrt{5}+2\right)^i}{2 \sqrt{5}}\right)=-\frac{\sqrt{5} \left(2-\sqrt{5}\right)^n-3 \left(2-\sqrt{5}\right)^n+\sqrt{5} \left(\sqrt{5}+2\right)^n+3 \left(\sqrt{5}+2\right)^n-2 \sqrt{5}}{2 \sqrt{5} \left(\sqrt{5}-1\right) \... 1 The syntax is mostly correct. The only improvement I see is to write the k-th derivative as f^{(k)}, not f^k. This also applies when k is a known number. There is an alternative notation with apostrophes. So you should write f^{(0)}(x) or f(x), f^{(1)}(x) or f'(x), f^{(2)}(x) or f''(x) and so forth. The second one is generally only used ... 3 Identify \Bbb R^2 with \Bbb C. Then (\cos x,\sin x) becomes \cos x+i\sin x=e^{ix} and so$$ g(\alpha)=\sum_{n=1}^\infty {(-1)^{n+1}}n^{-\alpha+it}$$Maybe that halps. 1 This is a second-order linear recurrence. The auxillary quadratic here is$$\lambda^2 = 4\lambda + 1,$$so you have$$\lambda^2 - 4\lambda - 1 = 0.$$Using the quadratic formula you get the bases$$\lambda = {4\pm \sqrt{16 + 4}\over 2} = 2\pm \sqrt{5}. $$Your solution is of the form$$E(n) = c_0(2 -\sqrt{5})^n + c_1(2 + \sqrt{5})^n.$$Match the initial ... 1 For conditionally convergence use the Leibniz convergence theorem, the sequence is decreasing and converges to zero. But the series does not converge absolute:$$\frac{\sqrt{k}}{k+1}\geq \frac{1}{k+1}$$which does diverge. 0 Hint. One may write, as k \to \infty, using a Taylor series expansion,$$ \begin{align} \frac{\sqrt{k}}{k+1}&=\frac1{\sqrt{k}}\frac1{1+\frac1{\sqrt{k}}} \\&=\frac1{\sqrt{k}}\left(1-\frac1{\sqrt{k}}+O\left(\frac1k\right)\right) \\&=\frac1{\sqrt{k}}-\frac1k+O\left(\frac1{k^{3/2}}\right) \end{align} $$thus the given series$$ \sum_{k\ge k_0}\frac{... 2 The reason for then+1$in$\displaystyle \int_1^{n+1} \frac{1}{x}\, dx \le \sum_{k=1}^n \frac{1}{k}$can be seen from this diagram from Wikipedia so$\displaystyle \int_1^{6} \frac{1}{x}\, dx \le \frac11+\frac12+\frac13+\frac14+\frac15$In effect they took$p=0, \, q=n$for the left hand inequality, possibly because it is more natural, and then they ... 2 The value of$p$is different for each of the inequalities. The inequality$\displaystyle \int_1^{n+1} \frac{1}{x}\, dx \le \sum_{k=1}^n \frac{1}{k}$comes from taking$p=0$and$q=n$in the theorem. The inequality$\displaystyle \sum_{k=2}^n \frac{1}{k} \le \int_1^n \frac{1}{x}, dx$comes from taking$p=1$and$q=n$in the proposition. (It should also be ... 0 for the second limit you can write $$n(\sqrt{n^2+2l}-n)=\frac{n(n^2+2l-n^2)}{\sqrt{n^2+2l}+n}=$$ can you proceed? 1 This is partial fraction decomposition. If you have $$\frac x {(x-a)(x-b)}=\frac A {x-a}+\frac B {x-b}$$ that is to say $$x=A(x-b)+B(x-a)=(A+B) x-(Ab+Ba)$$ Comparing terms, you then have equations $$1=A+B$$ $$0=Ab+Ba$$ Solve them for$A,B$to get $$\frac x {(x-a)(x-b)}=\frac{a}{(a-b) (x-a)}-\frac{b}{(a-b) (x-b)}$$ 0 $$\frac A{ax+1}+\frac B{bx+1}=\frac{(Ab+Ba)x+(A+B)}{(x+a)(x+b)}$$ and you are in a case such that $$A+B=0.$$ 1 Yes, you should add$\Delta t \to 0$to be explicit, but in general,$o(\Delta t)$is understood to hold for$\Delta t \to 0$, so I think it is fine not to add it, but there is no harm in writing it. To both notations you give: They talk about different things. The first one is the definition of$y$being differentiable at$t$, iff $$y(t + \Delta t) = y(... 3 Let x_n = {1 \over n} \lfloor nx \rfloor. Then x_n is rational for all n and x_n \to x for any x. There is nothing special about the sequence n, any rational sequence q_n such that q_n \to \infty will do, that is x_n = {1 \over q_n} \lfloor q_n x \rfloor. 0 1,1.4,1.41,1.414,1.4142.\ldots, converges to \sqrt2 In general, for any real number a=n.a_1a_2a_3a_4\ldots, where n is an integer and a_i is a decimal digit, then the sequence$$n, n.a_1, n.a_1a_2, n.a_1a_2a_3, \ldots$$converges to a. 0 The distance function d: (Y \times Y, d_{max}) \to \Bbb R is continuous. Hence, for all n \ge n_0, we have for all x \in X,$$d(f(x), f_n(x)) = d(\lim_{m \to \infty} f_m(x), f_n(x)) = d( \lim_{m \to \infty} (f_m(x), f_n(x))) = \lim_{m \to \infty} d(f_m(x), f_n(x)) \le \epsilon$$1 Now you can use induction to show that \displaystyle a_n=1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{(n-1)!}, so then \displaystyle\lim_{n\to\infty}a_n=\sum_{k=1}^{\infty}\frac{1}{k!}=\color{red}{e-1} 10 Instead, you should have got:$$a_{n+1} = a_n + \frac1{n!} = a_{n-1} + \frac1{(n-1)!} + \frac1{n!}= \cdots = a_1 + \frac1{1!} + \frac1{2!} + \cdots + \frac1{(n-1)!} + \frac1{n!}$$i.e.$$a_{n+1} = \sum_{k=1}^n \frac1{k!} = \sum_{k=0}^n \frac1{k!} - 1$$Hence, the limit is actually e - 1. If, instead, the sequence was defined as a_0 = 0 and a_1 =1,... 0 The usual power series expansion (1-x)^{-3} = \sum_{k\geq 0} \left( \begin{matrix} -3 \\ k \end{matrix} \right) (-x)^3 works (in fact for any complex number instead of -3). Here,$$ \left( \begin{matrix} -3 \\ k \end{matrix} \right) = \frac{(-3)(-4) \cdots (-3-k+1)}{k!}.$$Proof by Taylor expansion (as already mentioned). 4 Your mistake$$f'(0)=2 \quad \text{and} \quad f''(0)=4$$Another way to proceed: if you know the series expansion of both e^x and \sin(2x):$$e^x=1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots\sin(2x)=2x-\frac {4x^3}3+\cdots$$Then$$e^x\sin(2x)=\left(1+x+\frac {x^2}{2}+\frac {x^3}6+\cdots\right)\left(2x-\frac {4x^3}3+\cdots\right)=2x-\frac {4x^3}3+... 1 Based on your work above$f''(0) = 4$check this. Where did the$x^1$term go?$f(x) = \frac {0}{0!} + \frac {2}{1!}x + \frac {4}{2!}x^2 + \frac {-2}{3!} x^3 \cdots$I would say$3^{rd}$degree polynomial, and not "grade." But that would be correct, the highest exponent of$x = 3.$And, the highest derivative you would need is 3. What about the ... 1 You miscalculated the first and second order terms, remember that$sin(0) = 0$and$ cos(0) = 1$. 1 Assuming that the series converges absolutely, e.g. if all the elements are positive, you can always rearrange the series to be indexed by$\omega$, and get it over with. If that this is not the case, e.g. the first$\omega$elements are$-\frac1n$, you probably want to add the requirement that for every limit ordinal, the sum does converges. However, in ... 0 Let$(x_n)_{n\in N}$be a real sequence such that$\sum_{n=1}^{\infty} |x_n| <\infty$, We can re-index$(x_n)_{n\in N}$in any manner we like and still get the same value for the sum of the series. That is , let$f:N\to S$be a bijection. Let$y_{f(n)}=x_n$for$n\in N.$Let$A=\{s\in S:y_s\geq 0\}$and$B=\{s\in S: y_s<0\}.$Let$P$be the lub of ... 2 If$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-...

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