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5

There is no such series, and this is not something specific to power series. Claim If a sequence of continuous functions on $E$ converges uniformly on a dense subset of set $E$, then it converges uniformly on $E$. Proof. Uniform convergence is equivalent to being Cauchy in the uniform norm, which means $$ \forall \epsilon\ \exists N \text{ such that ...


3

It's not a power series but a Laurent series. You can express elements of $\mathbb{Q}_p$ as series $$\sum_{k=m}^\infty a_k p^k,$$ where $m\in\mathbb{Z}$ and each $a_k$ is an integer between $0$ and $p-1$ inclusive. For $\frac{1}{3!}$ in $\mathbb{Q}_3$, the shortcut way to find the series is to write $$\frac{1}{3!} = \frac{1}{3}\cdot \frac{1}{2} = ...


2

It is $$\frac{d^2}{dx^2}\left[\sum_{k=0}^\infty\frac{c_k}{k!}x^k\right]=\frac{d}{dx}\left[\sum_{k=1}^\infty\frac{k\cdot c_k}{k!}x^{k-1}\right]=\sum_{k=2}^\infty\frac{k(k-1)\cdot c_k}{k!}x^{k-2}.$$ Note that $$\frac{d^2}{dx^2}\left[c_0+c_1x+\sum_{k=2}^\infty\frac{c_k}{k!}x^k\right]=\frac{d}{dx}\left[c_1+\sum_{k=1}^\infty\frac{k\cdot ...


2

The "sophisticated" solution: $$f(x)=\sum_{n=1}^\infty a_nx^n$$ Here the sum is interpreted as the integral with respect to $\sum_{n=1}^\infty \delta_n$ of the function $$ F(n,x) = a_n x^n $$ Hence if we can apply the derivation theorem: $$ f'(x) = \frac d{dx} \sum_{n=1}^\infty F(n,x) \color{red}= \sum_{n=1}^\infty \frac d{dx}F(n,x) = \sum_{n=1}^\infty ...


2

You're right. This series defines function: $$e^{3x}=\sum_{n=0}^{\infty}\frac{3^nx^n}{n!}$$ So it's convergent everywhere.There's something wrong with answer $\frac{1}{3}$.


2

By the ratio test we get $$\frac{a_{n+1}}{a_n}=3\frac{\left(\frac23\right)^{n+1}+1}{\left(\frac23\right)^{n}+1}\xrightarrow{n\to\infty}3\implies R=\frac13$$ Remark Notice that if $\sum a_n x^n$ and $\sum b_n x^n$ are two power series with radius of convergence $R_a$ and $R_b$ and if $R_a\ne R_b$ then the radius of convergence of $\sum (a_n+b_n)x^n$ is ...


1

Thew ratio and root tests are for series of positive terms. The series is not absolutely convergent. You can use Dirichlet's test to prove that it is (conditionally) convergent.


1

You can do this somewhat directly: Look at $\frac{1}{4n} + \frac{i}{4n+1} + \frac{i^2}{4n+2} + \frac{i^3}{4n+3}.$ Since $i^4 = 1$, we know we can chunk up the series in this way. So, put all of these together as one fraction: you get $\frac{(16+16i)n^2 + (16+8i)n + 3}{4n(2n+1)(4n+1)(4n+3)}$. But this is a sum in which the terms are order $\frac{1}{n^2}$, ...


1

Yes, they are equal. @MisesEnForce showed the convergence. The equality follows from Abel's theorem on power series. http://en.wikipedia.org/wiki/Abel%27s_theorem and the continuity of $\log$. Example: $- \log 2 = \sum_{n \ge 1} \frac{(-1)^n}{n}$. Try for $z\ne 1$ a root of $1$. A conclusion from AbeI's theorem: if on at some points $z$ on the circle of ...


1

From looking at the first few $n$, it seems that in fact you don't need the sum: $$ \dfrac{1}{2\pi} \int_0^{2\pi} P_n(\cos \phi)\; d\phi = P_n(0)^2 $$ where in fact for odd $n$, both sides are $0$. The right sides are easy to find using Bonnet's recursion formula $$ (n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x) $$ I'm not sure how best to get the ...


1

Yes, assuming $|zw|<1$. $\mbox{ }$



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