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4

You can use l'Hospital, $$\lim_{x\to+\infty}\frac{\int_0^x e^{t^2}\,dt}{e^{x^2}/x}=\lim_{x\to+\infty}\frac{e^{x^2}}{2e^{x^2}-e^{x^2}/x^2}=\lim_{x\to+\infty}\frac{1}{2-1/x^2}=\frac{1}{2}.$$

1

Let me try. We have $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}.$$ Taking derivative two times, we have $$\sum_{n=2}^\infty n(n-1)x^{n-2} = \frac{2}{(1-x)^3}.$$ So, we have $$\sum_{n=2}^\infty n(n-1)x^{n-1} = \frac{2x}{(1-x)^3},$$ or $$\sum_{n=1}^\infty n(n+1)x^{n} = \frac{2x}{(1-x)^3}.$$ Substituting $x=\frac{1}{2}$, you get the result.

1

Hint: What is the Maclaurin series of $f(x) = \frac{1}{1-x}$ if $|x| < 1$? What does this tell you about the Maclaurin series of $f'(x)$, $f''(x)$, etc.?

0

Root test may save you: $$\lim_{n\to\infty} |n^n x^{n^2}|^{1/n} = \lim_{n\to\infty} n |x|^n = \begin{cases} \infty, & |x| \geq 1 \\ 0, & |x| < 1. \end{cases}$$ Therefore the series converges if and only if $|x| < 1$ and hence the radius of convergence is 1. If you want to apply the Cauchy-Hadamard theorem, notice that the coefficients of ...

2

Use Stirling's approximation $$n! \sim n^n e^{-n} \sqrt{2 \pi n} \quad (n \to \infty)$$ You should be able to conclude that the radius of convergence is $e$.

1

Here's a start: $$\left | e^x - \sum_{k=0}^{N-1} \frac{x^k}{k!} \right | \leq \sum_{k=N}^\infty \left | \frac{x^k}{k!} \right | \leq \sum_{k=N}^\infty \left | \frac{x^k}{N!N^{k-N}} \right | = \frac{1}{N! N^{-N}} \sum_{k=N}^\infty \left | \frac{x^k}{N^k} \right |.$$ Now the last sum is well approximated by $\frac{x^N}{N^N}$, if $N$ is sufficiently large. ...

2

It seems that the answer depends on the ring $R$. In the comments, many cases where an isomorphism cannot hold are given: If $R$ is a field, then $R[[x]]$ is local (i.e. it has only one maximal ideal), while $R[x]$ is not. If $R$ is non-trivial and finite or countable, then $R[x]$ is countable, while $R[[x]]$ is not. However, there are examples of ...

1

As per the comments: This is the same series as $\pi^j\sum\limits_{k=1}^n k^{2j}$.

0

The answer to the question is yes. This follows from Abel's theorem, since you assume the $a_k$ (and hence $k a_k$) are positive. If you assume that $\sum_k k a_k$ converges, then it is an immedeate consequence of Abel's theorem that $\sum_k k a_kx^{k-1}$ converges for $x\rightarrow 1^-$ to $\sum_k k a_k$. If, on the other hand, the sum does not converge, ...

4

No: consider $$f(x) = \log{(1+x^2)}.$$ Then $$f(x) = -\sum_{k=1}^{\infty} \frac{(-x^2)^k}{k},$$ which is an alternating series with decreasing terms for $-1 \leqslant x \leqslant 1$, so it converges on $[-1,1]$. It does not converge outside this interval by applying the ratio test. Then $$f'(x) = \frac{2x}{1+x^2} = \sum_{k=0}^{\infty} 2x(-x^2)^k$$ for ...

2

Your idea goes in the right direction, but doesn't quite work out. If we choose a branch of $\sqrt{z}$ on a domain where one exists, and expand $e^{\sqrt{z}}$, we get $$\sum_{n = 0}^\infty \frac{(\sqrt{z})^n}{n!} = \sum_{k = 0}^\infty \frac{(\sqrt{z})^{2k}}{(2k)!} + \sum_{k = 0}^\infty \frac{(\sqrt{z})^{2k+1}}{(2k+1)!} = \sum_{k = 0}^\infty ... 1 We can cheat and solve the ODE using separation of variables. The general solution for x>-1 is$$y(x)=c_0\sqrt{\mathstrut1+x}\ .$$Using the formula for the binomial series therefore gives$$y(x)=c_0\sum_{k=0}^\infty{1/2\choose k}x^k\ .$$Of course the numbers a_k:={1/2\choose k} can be written in terms of factorials, if desired. 0 Did you consider that perhaps it does not converge for some points on the boundary of the disk of convergence? Try z = \frac{i}{\sqrt{3}} so that the signs cancel out. And try to find other such points of divergence. \def\less{\smallsetminus} If you ignore the scalings, your function is just z \mapsto \ln(1+iz)-\ln(1-iz) within the region D[0,1] ... 0 @Andre Nicolas I got something totally differet. Separated as recommended for two series. and got: \frac{x}{1-x} - \frac{1}{x(1-x)^2} 0 Hint: Why not look at \sum \left(1-\frac{1}{n+1}\right)x^n? Two series, one very familiar, the other almost as familiar. Added: We have \sum_1^\infty x^n=\frac{x}{1-x}. Also, -\sum_1^\infty \frac{x^n}{n+1}=\frac{1}{x}\sum_1^\infty -\frac{x^{n+1}}{n+1}. Finally, \sum_1^\infty -\frac{x^{n+1}}{n+1}=\ln(1-x)+x. 2 You can check that$$ \left| \frac{z-\alpha}{1-\bar \alpha z} \right| < 1 $$precisely when |z|<1. (See for example this, but there are many many others on this site as well.) Hence the series converges for |z| < 1 (and diverges for |z| > 1 where \left| \frac{z-\alpha}{1-\bar \alpha z} \right| > 1). Finally, by Dirichlet's test, ... 1 There is a more general result here, and it has nothing to do with the irrationality of \pi. Claim: \sum \sin (nx)z^n diverges for all z,|z|=1 and for all x\in \mathbb {R}\setminus \pi\mathbb {Z}. I'll treat only the case 0< x \le 1. (See if you can fill in the details for the other values of x.) The proof is simple: As we view e^{inx} ... 0 Look at my question and resolution, this might help you with your exercise: Is series \displaystyle\sum^{\infty}_{n=1}\frac{\cos(nx)}{n^\alpha}, for \alpha>0, convergent? It's almost the same situation, apply it to your problem. 1 Hint: Write as$$\Im \left[ \sum_{n=1}^\infty (e^{i}z)^n \right].$$4 EDITED -1 \le \sin(n) \le 1, so the series converges for |z| < 1. Irrationality of \pi implies \sin(n) takes values arbitrarily close to \pm 1, so it diverges for |z| \ge 1. 4 As (at the time I was typing) no-one else had posted a complete answer, I'm reinstating mine, with corrections due to insightful comments from A.S. Because \pi is irrational, the additive subgroup of \Bbb{R} generated by \Bbb{Z} and \pi/2 is dense in \Bbb{R}. This means that for any N \in \Bbb{N} and \epsilon > 0 there is n > N such ... 1 First we write the power series in the following form:$$\sum_{n=1}^\infty\frac{(-1)^n}{n^2}z^{n(n+1)}=\sum_{k=1}^\infty a_kz^k$$then$$a_k=\begin{cases}\frac{(-1)^n}{n^2}, &\text{ if }k\text{ is of the form }n(n+1)\\ 0,&\text{otherwise}\end{cases}$$Then to find the radius of convergence we need to calculate ... 1 I just realized this question is trivial. The terms of the power series must go to zero, so for |z| < \delta there exists M > 0 such that |a_k z^k| \leq M. We apply this at z = \epsilon \delta for any \epsilon \in (0,1) and we obtain b = (\epsilon \delta)^{-1}. Done. 1 In fact, the partial sums are bounded. Note that for z with |z| = 1 and z \neq 1,$$ \left|\sum_{k=0}^n z^n\right| = \left|\frac{1 - z^{n+1}}{1 - z}\right| \leq \frac{|1 - z^{n+1}|}{|1 - z|} \leq \frac{2}{|1-z|} $$The Dirichlet criterion applies. 0 Remember that the radius of convergence is defined as the supremum of convergent series, so there is no need to look on the boundary point itself, if you can get away with a limiting argument. In this case, since it doesn't converge at z=1 we know R\le1, and conversely we can consider an arbitrary z with |z|<1; then ... 1 The most general criterion, which gives an exact formula for the radius in all cases, says that the radius of convergence of a power series is R iff \frac{1}{R} = \limsup |a_n|^{1/n}. This is known as Cauchy–Hadamard theorem. Since you assume that the radius of \sum a_n x^n is R, you know that \limsup |a_n|^{1/n} = \frac{1}{R}. Now you want to ... 5 If the power series converges at some a \neq 0, then it is absolutely convergent in |x| <|a|. In particular, it is continuous. Now as f(x) = \sum c_nx^n is zero on a dense subset of |x|<|a|, the continuity of f implies that f is identically zero in |x|<|a|. Now we argue that c_n = 0 for all n. First of all, put x = 0 into the ... 0 Use division of the numerator by the denominator along the increasing powers of x (not Euclidean division)! For an example, you can look at this thread. 0 Never forget you can do long division with polynomials 1 By definition, the the limit of the power series is the pointwise limit of the partial-sum polynomials in the range where the series converges (because that's what it means for the series to converge). However, the polynomials in the limit will not be the monomials c_nx^n you speak of, but entire prefixes of the power series:$$ c_0 \\ c_0+c_1 x \\ ...

11

It was shown in the answers to this question that $$e^{-x}\sum_{k=0}^n\frac{x^k}{k!} = \frac{1}{n!}\int_x^\infty e^{-t}\,t^n\,dt,$$ so setting $x=-n$ we have \begin{align} e^{n}\sum_{k=0}^n\frac{(-n)^k}{k!} &= \frac{1}{n!}\int_0^\infty e^{-t}\,t^n\,dt + \frac{1}{n!}\int_{-n}^0 e^{-t}\,t^n\,dt \\ &= 1 + \frac{(-1)^n}{n!} \int_0^n e^u u^n\,du ... 1 I only got asymptotic inequality. Numerical exploration tells us that the quantity in question - properly rescaled monotonically approaches the left limit in the inequality. Asymptotics. After expanding e^n=\sum_{k=0}^{n}\frac{n^k}{k!}+\frac1{n!}\int_0^ne^t(n-t)^ndt, making a substitution t=nx in the integral and rearranging the terms, the original ... 1 The answer to your question is no. First think about \int_0^z e^{-t^2} \, dt: $$it's an entire function, because it is the integral of an entire function, but infamously cannot be expressed in terms of elementary functions. You can then probably use some sort of differential Galois theory to prove that you can't build a finite set of primitives (likely ... 2 You have proved it for k=2. Let's suppose that it holds for k, we'll show it holds for k+1. Let q=(a+bi)^k. Note first that$$(a+ib)^{k+1}=q^k(a+bi)=aRe(q)-bIm(q)+(aIm(q)+bRe(q))i $$Now: A^{k+1} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}^{k+1}=\begin{bmatrix} a & -b \\ b & a \end{bmatrix}^{k}\begin{bmatrix} a & -b \\ b ... 2 Hint: Convergence is not necessary; consider \limsup. 1 Hint: The easier way is to prove that$$J=\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$$has the property J^2=-I and you are trying to prove that A=aI+bJ is "like" a+bi. Then the induction step is much cleaner. 2 There is a field isomorphism \eta :\Bbb C\to F where F is the set of matrices M(a,b)= \begin{pmatrix} a&b\\-b&a\end{pmatrix},a,b\in\Bbb R under usual matrix sum and multiplication that sends a+bi to M(a,b) and your claim follows from this. 1 There is really no problem with your 3 examples since they are totally different representations which don't have to have the same convergence radius. An easier example to investigate this issue would be the following: Let f(x)=\frac{1}{x}. This function is defined for all non-zero x. Now, by the famous geometric series, one could write this as ... 2 A re-worked answer. We have:$$ e^n = \sum_{k=0}^{n}\frac{n^k}{k!}+\frac{n^{n+1}}{n!}\int_{0}^{1}\left(e^t(1-t)\right)^n\,dt\tag{1}$$but we also have:$$\forall x\in[0,1],\qquad (1-x^3)e^{-x^2/2} \leq e^{x}(1-x) \leq e^{-x^2/2}\tag{2}$$hence:$$ \int_{0}^{1}\left(e^t(1-t)\right)^n\,dt \leq \int_{0}^{+\infty}e^{-nx^2/2}\,dx = \sqrt{\frac{\pi}{2n}}\tag{3}$$... 0 y′=\sum_{n=-1}^{\infty}a_{n+1}(n+1)x^n y′'=\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^n \lambda y = 2xy'-(1-x^2)y'' \lambda y = 2x[\sum_{n=-1}^{\infty}a_{n+1}(n+1)x^n]-(1-x^2)[\sum_{n=-2}^{\infty}a_{n+2}(n+2)(n+1)x^n] \lambda y = ... 0 You could certainly use power series, but I use the expansion about the point x=-3 rather than x=0 (since you have those (x+3) terms). Alternatively, let z(x)=(x+3)y(x), so z'=(x+3)y'+y and z''=(x+3)y''+2y'. You can then find a fairly simple ODE for z and solve it however you like. 1 There is a life hack just for you: solve the thing with Wolfram, see how it could be simplified, then solve it the way you were supposed to. What if we switch to u(x)=(x+3)\cdot y(x) and look for that in the form of power series? 7 Assume neither A=0 nor B=0. Let n be minimal with a_n\ne 0, let m be minimal with b_m\ne 0. Then the coefficient of x^{n+m} in the Cauchy product equals a_nb_m\ne 0. 2 Since \frac1{1-x} =\sum_{n=0}^{\infty} x^n , we have x^m\frac1{1-x} =\sum_{n=0}^{\infty} x^mx^n =\sum_{n=0}^{\infty} x^{m+n} =\sum_{n=m}^{\infty} x^{n} . Subtracting, \sum_{n=0}^{m-1} x^{n} =\sum_{n=0}^{\infty} x^{n}-\sum_{n=m}^{\infty} x^{n} =\frac1{1-x}-\frac{x^m}{1-x} =\frac{1-x^m}{1-x} . A nice thing about this result is that it is true for any ... 1 Hint: can you do$$ \sum_{k=15}^{\infty} (-3)^kx^{2k}? $$2 Another way: 1+x+x^2+\cdots+x^n=\frac{x^{n+1}-1}{x-1}. Take derivative 1+2x+\cdots+nx^{n-1}=\frac{(n+1)x^{n}(x-1)-(x^{n+1}-1)}{(x-1)^2}. Since a is n-th root of unity. We get \frac{n}{a-1}. 3 Hint: It sounds like you may have made an algebraic slip when computing (1-a)S (don't worry, it happens to all of us every now and then!). Try computing it again and you should get a geometric series plus a negative term. Remember that a^n = 1 and try substituting that into what you have. It should fall out pretty easily. Hope this helps! 1 What is the absolute value of n'th term? Check it. You might be surprised. 0 Since:$$ U(a,b,x)=\frac{1}{\Gamma(a)}\int_{0}^{+\infty}e^{-xt}t^{a-1}(1+t)^{b-a-1}\,dt\tag{1}$$we have:$$\begin{eqnarray*} \frac{d}{dx}U(a,b,x) &=& -\frac{1}{\Gamma(a)}\int_{0}^{+\infty}e^{-xt}t^{a}(1+t)^{b-a-1}\,dt\\&=& -a\cdot U(a+1,b+1,x).\tag{2}\end{eqnarray*}$$2 First note if f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)z^k}{k!}, then$$\begin{align} \oint_{|z|=1}f(z)\,dz&=\oint_{|z|=1}\sum_{k=0}^{\infty}\frac{f^{(k)}(0)z^k}{k!}\,dz\\\\ &=\sum_{k=0}^{\infty}\frac{f^{(k)}(0)}{k!}\oint_{|z|=1}z^k\,dz\\\\ &=0 \end{align} since $\oint_{|z|=1}z^n\,dz=0$ for all $n\ne -1$ (as written in the OP). However, ...

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