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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 $k$ time 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 $n$th 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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