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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 ...

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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 ...

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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 $|z| \ge 1$.

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

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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, ...

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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 ...

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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 $e$.

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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\ .$$Of course the numbers a_k:={1/2\choose k} can be written in terms of factorials, if desired. 1 Hint: Write as$$\Im \left[ \sum_{n=1}^\infty (e^{i}z)^n \right].$$1 There is a more general result here, and it has nothing to do with the irrationality of \pi. Claim: \sum \sin (nx)z^n diverges for all z,|z|=1 and for all x\in \mathbb {R}\setminus \pi\mathbb {Z}. I'll treat only the case 0< x \le 1. (See if you can fill in the details for the other values of x.) The proof is simple: As we view e^{inx} ... 1 As per the comments: This is the same series as \pi^j\sum\limits_{k=1}^n k^{2j}. 1 Here's a start:$$\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 |.$$Now the last sum is well approximated by \frac{x^N}{N^N}, if N is sufficiently large. ... 1 Hint: What is the Maclaurin series of f(x) = \frac{1}{1-x} if |x| < 1? What does this tell you about the Maclaurin series of f'(x), f''(x), etc.? 1 Let me try. We have$$\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.

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