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2

Hint: $$\frac{n^3}{n!}=\frac{n^2}{(n-1)!}=\frac{1}{(n-1)!}+\frac{n+1}{(n-2)!}=\frac{1}{(n-1)!}+\frac{3}{(n-2)!}+\frac{1}{(n-3)!}$$

1

The relative error for $\pi^k$ after summing $n$ terms is $\approx n^{-k}$. Computing the $k$th root then does little change (the relative error becomes $\frac 1kn^{-k}$). Hence the number of correct digits is essentially $k\log_{10}n$. For any fixed $k$, this does not grow very well if we compare it to what the Borweins managed (in the linked Wikipedia ...

1

Write $f(z)=\sum_n a_nz^n$ and $g(z)= {{f(z)}\over{z^n}}$, $g(z)=a_0/z^n+...+a_{n-1}/z+\sum_{l\geq n}a_lz^{l-n}$. Since $\mid g(z)\mid <M$ for $\mid z\mid >r$, we deduce that $\mid \sum_{l\geq n}z^{l-n}\mid-\mid a_0/z^n+...a_{n-1}/z\mid<M$. This implies that the entire function $h(z)=\sum_{l\geq n}a_nz^{l-n}$ is bounded so it is a constant. Thus ...

1

Aha! Thanks to Piotr's tip (and the now-corrected summation error in the $y'$ expression), if I take the second-to-last equation $$\sum_{n=0}^{\infty} \left( \frac{2n \cdot x^{2n}-x^{2n+2}}{2^nn!}\right)=0$$ I can rewrite it as $$\sum_{n=0}^{\infty} \frac{2nx^{2n}}{2^nn!}-\sum_{n=0}^{\infty} \frac{x^{2n+2}}{2^nn!} =0 \quad \Rightarrow \quad ... 2 By inspection, y=\exp\tfrac{x^2}{2} so$$y'=xy,\,y''=\left(1+x^2\right)y,\,y''-xy'-y=y\left(1+x^2-x\cdot x-1\right)=0.$$2 You wrote y'=...=x+x^3/2+x^5/8+...=-1+\sum_{n=0}^{\infty}x^{2 n+1}/2^n n! which is a mistake because the "-1" should be erased. With that correction,there is no "x" term in your last line and the summation is 0 because it is a telescoping series of the form (a_0-a_1)+(a_1-a_2)+(a_2-a_3)+.... with a_0=0 and with a_n converging to 0. You can ... 2 The trick here is to compare the terms in front of x of the same power. So rewrite the sum with x^{2n+2} term to a x^{2n} (just shift all the n's in the sum by one. You can do that because different sums are independent of each other.) and then compare. 1 I think this is true because, in a neighbourhood of 1 (1-\epsilon , 1)  you have that \ a_k\ x^k \geq 0 \ \ \forall k so you can interchange the limits (The infinite sum is a limit),so:$$b = \lim\limits_{N\rightarrow +\infty} \lim\limits_{x\rightarrow1^-} \sum\limits_{k=0}^N a_kx^k =\lim\limits_{N\rightarrow +\infty} \sum\limits_{k=0}^N a_k \geq ...

2

Wlog we may assume that $x > 0$. For any $n \in \mathbb{N}$ we get $$\sum_{k=0}^{n}a_k x^{k} \leq \sum_{k=0}^{\infty} a_k x^{k}$$ by assumption. Taking the limit of $x \to 1$ in this inequality shows the claim.

1

What you've stated is the converse to Abel's theorem, which isn't true in general. That is, it's not the case that $\sum_{k=0}^\infty \, a_k$ converges. A paper which gives conditions under which a converse does hold is "The converse to Abel's theorem on power series" by H. Delange in Annals of Math. 50 No.1 (1949)

1

If you insist on using the Hadamard formula, here's how it can be done. $$\sum_{n=0}^\infty (-1)^n 2^n z^{2n+2} = z^2 - 2z^4 + 4z^6 - 8z^8 + 16z^{10} - \dotsb := \sum_{n=2}^\infty b_n z^n$$ where $b_n = 0$ if $n$ is odd and $(-2)^{n/2-1}$ if $n$ is even. Then the sequence $|b_n|^{1/n}$ alternates between $0$ and $2^{1/2-1/n}$ according to the parity of ...

2

Yes, you went wrong because of the $z^{2n}$. You're assuming that $a_n=(-1)^n2^n$. That's not so. In fact $a_{2n+2}=(-1)^n2^n$, while $a_k=0$ if $k$ is not of the form $k=2n+2$. The best way to look at these things, in my opinion, is to forget that $|a_n|^{1/n}$ and know this: The radius of convergence is the supremum of the $r$ such that $|a_n|r^n$ is ...

0

Note that the given regions do not contain the singularities. A Laurent series is meaningful as long as the regions of definition do not contain the singularities. If the annuli are centered at a singularity the annuli will not contain the singularity.

1

The related theorem is that if a power series $$\sum a_k(z-z_0)^k$$ converges for some $z=z_1$ then it also converges for all $z$ with $$|z-z_0|<|z_1-z_0|.$$ Here $z_0=3$, $z_1=-1.1$ so that convergence is guaranteed for $|z-3|<4.1$. $z=7$ satisfies this inequality. Proof idea: The terms of the $z_1$ series converge to zero, thus are bounded by ...

2

Your reasoning is not quite right because you did not solve the compound inequality correctly and you did not actually prove that $|x-3|<r$ for $x=7$. Let's say we have the following, as you did: $$-r+3 < x < r+3$$ Then, we need to solve this compound inequality for $x=-1.1$. If $-r+3 < -1.1$, then $-r < -4.1$ and $r > 4.1$, or $4.1 ... 0 I don't think L'Hopital is a good way to go here. I would use i)$1-\cos x^2 = x^4/2 + O(x^8),$and ii) the power series defines a function on$(-1,1)$that has the form$4^5x^4 + x^5g(x),$where$g$is continuous at$0.$2 HINT: Recall that we have$1-\cos(x^2)=\frac12 x^4(1+O(x^4))$so that $$\frac{1}{1-\cos(x^2)}=\frac{2}{x^4}+O(1)$$ SPOILER ALERT Scroll over the highlighted area to reveal the solution 1 Say you have a coin which lands heads up with probability$p$, and you keep flipping it until it lands tails-up$k+1$times, recording the number of flips this takes as$X$. Then: $$\mathbb{P}(X=m+1)=\mathbb{P}(k \text{ tails in the first } m \text{ flips, then another tails})=\binom{m}{k}(1-p)^{k+1}p^{m-k}$$ Given that$k+1\le X < \infty$almost ... 5 Use the fact that for$|x|<1, \:k\geqslant0$$$\frac1{(1-x)^{k+1}}=\sum_{m=k}^{+\infty}\binom{m}{k}x^{m-k}$$ which can be proved by differentiating$ktime on both sides of following $$\frac1{1-x}=\sum_{m=0}^{+\infty}x^m$$ Thus \begin{align} \sum_{m=k}^{+\infty}\binom{m}{k}(1-p)^k\cdot p^{m-k}&=(1-p)^k\sum_{m=k}^{+\infty}\binom{m}{k}p^{m-k} \\ ... 2 In this ansswer, it is shown that \begin{align} \binom{m}{k} &=\binom{m}{m-k}\\ &=(-1)^{m-k}\binom{-k-1}{m-k} \end{align} Plug this into \begin{align} \sum_{m=k}^\infty\binom{m}{k}(1-p)^kp^{m-k} &=\sum_{m=k}^\infty(-1)^{m-k}\binom{-k-1}{m-k}(1-p)^kp^{m-k}\\ &=\sum_{m=0}^\infty(-1)^m\binom{-k-1}{m}(1-p)^kp^m\\ ... 1 And the "Bertrand series" idea continues indefinitely: \sum {1\over n(\log n)(\log\log n)^{1+\epsilon}} converges for \epsilon>0, and diverges for \epsilon=0. Similarly with as many-fold iterated log as you'd want. All from the integral test. 4 This is a Bertrand's series which are series of the form\sum_{k\ge 2}\frac1{k^\alpha(\log k)^\beta},$$and they're known to converge if and only if \alpha >1 (by the comparison test), or \alpha=1 and \beta>1 (by the integral test). 2 Root test:$$\sqrt[n]{|(3+\cos n)x^n|}\to |x|$$because$$2\le(3+\cos n)\le 4\implies \sqrt[n]{2}\le\sqrt[n]{(3+\cos n)}\le\sqrt[n]{4}.$$So the radius is 1. 1 Every convergent power series is infinitely differentiable in the same interval of convergence.$$ f'(x)=\sum_{n=1}^{\infty}na_nx^{n-1} $$So radius of convergence of f'(x) is$$ {1\over\limsup(n|a_n|)^{1\over n}}={1\over\limsup |a_n|^{1\over n}} $$as$$\lim_{n\to\infty}n^{1\over n}=1$$0 If we take the sum from n=0 on, then the exponent on the nth term is 3n+1 as you said. I'd just "construct" the coefficient. Start with a power of -1 to get the alternating behavior. Since we start low with n=0, (-1)^{n+1} gives us -1, +1, -1, +1, ... If we multiply this by \frac{1}{2} to decrease the difference from 2 to 1 (what we ... 1 The coefficients are alternating between 2 and 3, so you can do something like$$c_k = \frac{5+(-1)^k}{2}$$Now you just have to express k in terms of n, I think k=3n+1 might work. We obviously only care whether k is even or odd. So if n is even, k will be odd and vice versa. So we could also choose k=n+1. 2$$K(x)=4\sum_{n=0}^\infty \frac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)!}$$Differentiate two times to get$$K''(x)=4\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n+1)!}$$Multiply by x both sides:$$xK''(x)=4\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}=\sin(x)K''(x)=4\frac{\sin(x)}{x}$$Integrate two times from 0 to x:$$K'(x)=4\mbox{Si}(x)$$... 0 The general expression is \frac{K}{4a^2} = \sum_{k=0}^\infty (-1)^k\frac{a^{2k}}{(2k+1) \cdot (2k)!} Hint: Note that \int (-1)^k\frac{a^{2k}}{(2k+1) \cdot (2k)!} da = (-1)^k\frac{a^{2k+1}}{ (2k)!} 2 If you write down the expansion for \cos a, subtract the first term and then divide by a^2 you arrive at a power series is the term-wise derivative of (a linear multiple of) this series. So at least the derivative of your series, divided by a, with respect to a has a closed form. I am not sure if this is useful as a general approach. It is certainly ... 1 You can check the book "Singularities of Plane Curves" by Eduardo Casas-Alvero. If I remember correctly the first one or two chapters contain an elementary introduction to Newton-Puiseux series and the Newton method. 1 If a is an non-negative integer, the two solutions are exactly Legendre Polynomials, whose radius of convergence is +\infty. If a is a real number, the power series solution (using Frobenius Method) can be written in terms of the hypergeometric function,$$ x(t) = {}_2F_1\left(-a,a+1;1;\frac{1-t}{2}\right). $$Since the hypergeometric function ... 1 Defining$$a_n=\frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!}$$you can get rid of the numerator if you notice that$$\prod_{k=1}^{n-1} (2k-1)= \frac{2^{n-1} \Gamma \left(n-\frac{1}{2}\right)}{\sqrt{\pi }}$$which makes$$a_n=\frac{\Gamma \left(n-\frac{1}{2}\right)}{2 \sqrt{\pi } n!}$$which makes$$\frac{a_{n+1}}{a_n}=1-\frac{3}{2 (n+1)}$$As Brian M. Scott ... 1 To show convergence, first note that$$\prod_{k=1}^n(2k-1)=(2n-3)!!=\frac{(2n-2)!}{2^{n-1}(n-1)!}\;,$$so$$\frac{(2n-3)!!}{2^nn!}=\frac{(2n-2)!}{2^{2n-1}n!(n-1)!}=\frac1{2^{2n-1}n}\binom{2n-2}{n-1}\;.\tag{1}$$The ratio test doesn’t help here, but you can use the fact that the central binomial coefficient \binom{2n}n is asymptotically ... 0$$\sum_{k=1}^{n-1}(2k-1)=\dfrac{(n-1)}\cdot2(1+2n-3)=(n-1)^2=n(n-1)-n+1$$Now$$\dfrac{\sum_{k=1}^{n-1}(2k-1)}{2^n n!}=\dfrac{n(n-1)-n+1}{2^n n!}=\dfrac14\cdot\dfrac{(1/2)^{n-2}}{(n-2)!}-\dfrac12\cdot\dfrac{(1/2)^{n-1}}{(n-1)!}+\dfrac{(1/2)^n}{n!}$$Finally use e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!} 1 Here are hints to show that the series converges. (Actually computing the value will take more work.) Hint1: The series is \frac1{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10} + \cdots. Hint2: One attempt is to cancel the 1 with the 2, the 3 with the ... 4 The Taylor series for f(x) = e^{-x^2}, centered at x = 0, is indeed \displaystyle\sum_{n = 0}^{\infty}\dfrac{(-x^2)^n}{n!}. However, to evaluate f^{(39)}(0), you need to look at the coefficient of the x^{39} term (and then multiply that coefficient by 39!). The term \dfrac{(-x^2)^{39}}{39!} = -\dfrac{x^{78}}{39!} is not the correct term to ... 0 The radius of convergence r is at least \sqrt {17} and at most \sqrt {65}. We have convergence in (a) because |1+i|=\sqrt 2<\sqrt {17}< r and divergence in (b) because |9|=9>\sqrt {65}>r. We do not have sufficient information to decide (c). 3 There is an elementary path:$$I = \frac{1}{2\pi}\int_0^{2\pi}\frac{\sin^2\theta}{(1+\epsilon\cos\theta)^2}\,d\theta =\frac{1}{2\pi\epsilon}\int_0^{2\pi}\sin\theta\,d\left(\dfrac{1}{1+\epsilon\cos\theta}\right)$$By parts:$$ I = \frac{1}{2\pi\epsilon}\dfrac{\sin\theta}{1+\epsilon\cos\theta}\,\biggr|_0^{2\pi} - ... 1 To understand the first line, note that the Taylor series expansion of\frac 1 {(1 + a \epsilon)^2}$around$0$gives $$\frac 1 {(1 + a \epsilon)^2} = \sum \limits _{n=0} ^\infty (-1)^n (n+1) a^n \epsilon ^n .$$ In this, take$a = \cos \theta$and multiply the whole equality by$\sin ^2 \theta$(note that inside the parantheses in the right-hand side ... 0 Note that$\cos^3x=\frac{1}{4}(3\cos(x)+\cos(3x))$. But$\cos(x)=\sum_{k=0}^\infty\dfrac{(-1)^kx^{2k}}{(2k)!}\cos(3x)=\sum_{k=0}^\infty\dfrac{(-9)^kx^{2k}}{(2k)!}$Then$\begin{eqnarray} 4\cos^3(x)&=&3\cos(x)+\cos(3x)\\ &=&3\sum_{k=0}^\infty\dfrac{(-1)^kx^{2k}}{(2k)!}+\sum_{k=0}^\infty\dfrac{(-9)^kx^{2k}}{(2k)!}\\ ...

0

By using the identity $\cos(3x)=4\cos^3 x-3\cos x$ it follows \begin{align} \cos^3x&=\frac{1}{4}\cos (3x)+\frac{3}{4}\cos x\\ &=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\frac{(3x)^{2k}}{(2k)!}+\frac{3}{4}\sum_{k=0}^{\infty}(-1)^k\frac{(x)^{2k}}{(2k)!}\\ &=\frac{1}{4}\sum_{k=0}^{\infty}(-1)^k\frac{3(3^{2k-1}+1)}{(2k)!}x^{2k} \end{align}

0

Hint. One may recall that $$\cos^3 x=\frac14 \cos (3x)+\frac34\cos x$$ then use $$\cos x=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}, \quad x \in \mathbb{R}.$$

1

suppose i make a change of variable $t-1 = s,\space t = s+1$ and denote the derivative with respect to $s$ by $\cdot.$ we have the regular equation $$\ddot x+(s+1)\dot x+\frac1{3+3s+s^2} x = 0.$$ now look for solutions in the form $$x = a_0 + a_1s + a_2s^2 + a_3s^3+\cdots$$ we have (3+3s+s^2)\left(1 \cdot 2 a_2+2\cdot3a_3 s+3\cdot 4 ... 0 As \;\biggl(\dfrac1{1-x}\biggr)^{(n-1)}=\dfrac{(n-1)!}{(1-x)^n},\; you can derive term by term the power series expansion of \;\dfrac1{1-x}=1+x+x^2+\dots+x^m+\dotsm You obtain \begin{align*} \frac{(n-1)!}{(1-x)^n}&=\sum_{m\ge n-1}m(m-1)\dotsm(m-n+2)x^{m-n+1}\\ &=\sum_{m\ge n-1}\frac{m!}{(m-n+1)!}x^{m-n+1} =\sum_{m\ge 0}\frac{(m+n-1)!}{m!}x^m,\\ ... 0 There's a simpler version of the above formula:\frac{1}{(1-x)^n}=\sum_{k=0}^\infty \binom{k+n-1}{n-1}x^k$$You can prove this by induction - differentiate and then divide by n. 2 Yes its the binomial expansion for any index. (1-x)^{-n} = (-x)^{0} + -n(-x)^{1}+ \dfrac{-n(-n-1)}{2!}(-x)^{2} + ... which simplifies to .. (1-x)^{-n} = 1 + nx+ \dfrac{n(n+1)}{2!}(x)^{2} + \dfrac{n(n+1)(n+2)}{3!}(x)^{3} ... ie, (1-x)^{-n} = 1 + nx+ {n+1\choose 2}(x)^{2} + {n+2\choose 3}(x)^{3} ... Binomial expansion for any index is ... 2 You can observe that if |x|\leq 1, then$$ \sum_{n=1}^\infty \frac{|x|^n}{n(n+1)} \leq \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}, $$and thus the original series converges for all x\in [-1,1]. On the other hand, it is not necessary to know what is the value of \sum\frac{1}{n^2}, but only that \sum \frac{1}{n^\alpha} converges if and only if ... 1$$\lim_{n\to\infty}\dfrac{\dfrac{x^{n+1}}{(n+1)(n+2)}}{\dfrac{x^n}{n(n+1)}}=x$$So using d'Alembert's ratio test the series converges if |x|<1 and diverges if \cdots Now for x=1$$\sum_{n\to\infty}\dfrac1{n(n+1)}$$which is a p series with p=2 If f(x,n)=\dfrac{x^n}{n(n+1)}$$\sum_{n\to\infty}f(x,n)>\sum_{n\to\infty}f(-x,n)$$So, the ... 0$$\dfrac{x^n}{n(n+1)}=\dfrac{x^n}n-\dfrac1x\cdot\dfrac{x^{n+1}}{n+1}$$See Taylor series for \log(1+x) and its convergence 0 you can use the tree poles to calculate your series -1,0,1 XE around the zero$$\frac{x^2}{2 (x+1)}+\frac{x^2 \cos (2)}{2 (x-1)}-\frac{1}{2} x \cos (1)-\frac{\cos (1)}{x}+\sin (1)$$or around the x=1$$\frac{x^2}{8 (x+1)}+\frac{3 x^2 \sin (2)}{4 (x-1)}+\frac{5 x^2 \cos (2)}{8 (x-1)}-\frac{x}{4 (x+1)}+\frac{1}{8 (x+1)}-\frac{2 x \sin (2)}{x-1}+\frac{5 \sin ...

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