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5

Let's make it look nice. ∑[(x^(2n+1))/((x^2+1)^(n+1))]*[(2n!!)/(2n+1)!!] You say $\arctan(x) =\sum\dfrac{x^{2n+1}}{(x^2+1)^{n+1}}\dfrac{(2n)!!}{(2n+1)!!} $ Since $(2n+1)!! =\prod_{k=1}^n (2k+1) =\dfrac{\prod_{k=1}^n (2k)(2k+1)}{\prod_{k=1}^n (2k)} =\dfrac{(2n+1)!}{2^nn!} $ and $(2n)!!=2^nn! $, this becomes $\arctan(x) =\sum\dfrac{x^{2n+1}}{(x^2+1)^{n+1}}...


4

This is basically just a guess, but I would expect the limit to be $\infty$. Since $n \to \infty$, $\frac 1 n \to 0$, so we can ignore that term, giving us: $$\lim_{n \to \infty} \frac{n^n}{n^n} t^n=\lim_{n \to \infty} t^n=\infty$$


2

$$\arctan(x) = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}x^{2n+1} $$ for any $x\in(-1,1)$, so: $$ \frac{1}{10}-\frac{1}{3000}<\arctan\left(\frac{1}{10}\right) < \frac{1}{10}-\frac{1}{3000}+\frac{1}{500000}$$ and we may take $k$ as: $$ k = \left\lfloor \frac{100}{3}\cdot 299\right\rfloor = \color{red}{996}.$$


2

Being a finite sum, this one converges for every $x \in \Bbb C$, so its radius of convergence is $\infty$.


2

Because the $B(w_i,r_{w_i})$'s cover $frB(0,R)$, the $B(w_i,r_{w_i})$'s and $B(0,R)$ cover the clausure of $B(0,R)$. This is a open covering so it also covers $B(0,R+\varepsilon)$ for some $\varepsilon>0$. Now you have by definition that the radius of convergence of $F$ is $S>R$. This leads to a contradiction with the Cauchy-Hadamard formula because ...


2

Where did you find that equation? It's quite different from what I got, which I shall explain now. First, a common power series is $$ \frac{1}{1-x} = \sum_{i\geq0} x^{i}. $$ Using the substitution $x=-t^{n}$, $$ \frac{1}{1+t^{n}} = \sum_{i\geq0} (-t^{n})^{i} = \sum_{i\geq0} (-1)^{i}t^{in}. $$ Then, $$ \int_{0}^{x} \frac{1}{1+t^{n}} dt = \int_{0}^{x} \...


1

You're on the right track, but a few things need to be mentioned. First, if the series involves complex numbers then you're actually looking for the disk of convergence, not the interval. Second, your notation is very inappropriate. When you apply the root test, you're applying it to $a_n$ only. Including the summation symbol is incorrect. Third, your ...



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