E. K. K: Fill & Download for Free

Let [math]a_k=\prod_{j=1}^k\frac{j!\,e^j}{j^{j+0.5}}[/math]. The radius of convergence of a power series is typically found using either the root test or the ratio test; that is, computing either[math]\lim_{k\to\infty} \sqrt[k]{\left|a_k\right|}\qquad\text{or}\qquad\lim_{k\to\infty} \left|\frac{a_{k+1}}{a_k}\right|\tag*{}[/math]In this case, the ratio test seems more promising because [math]a_{k+1}[/math] and [math]a_k[/math] have a lot of common factors:[math]\dfrac{a_{k+1}}{a_k}=\dfrac{(k+1)!\, e^{k+1}}{(k+1)^{k+1.5}} =\dfrac{k!\, e^{k+1}}{(k+1)^{k+0.5}}\tag*{}[/math]According to Stirling's approximation, [math]\lim_{k\to\infty}\frac{k!\,e^k}{k^{k+0.5}}=\sqrt{2\pi}[/math]. Furthermore, [math]\lim_{k\to\infty}\left(\frac{k+1}{k}\right)^k=e[/math]. Therefore,[math]\begin{align*} \lim_{k\to\infty}\frac{a_{k+1}}{a_k} &=\lim_{k\to\infty}\frac{k!\, e^{k+1}}{(k+1)^{k+0.5}} \\ &=e\lim_{k\to\infty}\frac{k!\,e^k}{k^{k+0.5}}\cdot \frac{k^{k+0.5}}{(k+1)^{k+0.5}}\\ &=e\lim_{k\to\infty}\frac{k!\,e^k}{k^{k+0.5}}\cdot\lim_{k\to\infty} \left(\frac{k+1}{k}\right)^{-k}\cdot \lim_{k\to\infty}\sqrt{\frac{k}{k+1}}\\[5pt] &=e\cdot\sqrt{2\pi}\cdot e^{-1}\cdot 1\\[5pt] &=\sqrt{2\pi} \end{align*}\tag*{}[/math]Therefore, the terms of the series behave asymptotically like [math]\left(\sqrt{2\pi}x\right)^n[/math], so it converges when [math]|x|<\frac1{\sqrt{2\pi}}[/math].How might I prove that the radius of convergence of [math]\displaystyle \sum_{k=1}^\infty(\prod_{j=1}^k\frac{j!\times e^j}{j^{j+0.5}})x^k[/math] is [math]\frac{1}{\sqrt{2\pi}}[/math]?

We want to evaluate the integral[math]I = \displaystyle \int_0^{\infty} \lfloor x \rfloor e^{-x} \, dx. \tag*{}[/math]When seeing the greatest integer/floor function inside of an integral, split the integral accordingly where the floor function takes a fixed constant value:[math]\begin{align*} I &= \displaystyle \sum_{k=0}^{\infty} \int_k^{k+1} \lfloor x \rfloor e^{-x} \, dx\\ &= \displaystyle \sum_{k=0}^{\infty} \int_k^{k+1} ke^{-x} \, dx\\ &= \displaystyle \sum_{k=0}^{\infty} k(e^{-k} - e^{-(k+1)}). \end{align*} \tag*{}[/math]Next, we rewrite this as follows:[math]\begin{align*} I &= \displaystyle \sum_{k=0}^{\infty} (ke^{-k} - (k+1)e^{-(k+1)}) + \sum_{k=0}^{\infty} e^{-(k+1)}. \end{align*} \tag*{}[/math]The first term is a telescoping series[math]\displaystyle \sum_{k=0}^{\infty} (ke^{-k} - (k+1)e^{-(k+1)}) = \lim_{N \to \infty} (0e^0 - (N+1)e^{-(N+1)}) = 0, \tag*{}[/math]while the second term is a geometric series:[math]\displaystyle \sum_{k=0}^{\infty} e^{-(k+1)} = \frac{e^{-1}}{1 - e^{-1}} = \frac{1}{e - 1}. \tag*{}[/math]Hence, we conclude that[math]I = \displaystyle \int_0^{\infty} \lfloor x \rfloor e^{-x} \, dx = \frac{1}{e - 1}. \tag*{}[/math]

We are required to prove that [math](n!)^{1/n}>\frac{n}{3}[/math] by induction on [math]n[/math]. We rewrite this relation as [math]n!>\left(\frac{n}{3}\right)^n[/math], or [math]n^n<3^n\,n![/math]. (Note that this manipulation is permissible since [math]n[/math] is known to be positive.)We thus proceed to prove this statement [math]n^n<3^n\,n![/math] by induction on [math]n[/math].When [math]n=1[/math], the left hand side is [math]1^1=1[/math], while the right hand side is [math]3^1\,1!=3[/math]. Since [math]1<3[/math], the base case is proved.We now assume the inductive hypothesis, that is, [math]k^k<3^k\,k![/math] for some positive integer [math]k[/math][math][/math]. We are required to prove that [math](k+1)^{k+1}<3^{k+1}\,(k+1)![/math].The left hand side is [math](k+1)^{k+1}[/math]. By the binomial theorem, this expands to[math]k^{k+1}+(k+1)k^k+\dfrac{(k+1)k}{2!}k^{k-1}+\dfrac{(k+1)k(k-1)}{3!}k^{k-2}+\dfrac{(k+1)k(k-1)(k-2)}{4!}k^{k-3}+\cdots+1[/math][math]<k^{k+1}+(k+1)k^k+\dfrac{(k+1)k}{2!}k^{k-1}+\dfrac{(k+1)kk}{3!}k^{k-2}+\dfrac{(k+1)kkk}{4!}k^{k-3}+\cdots+\dfrac{(k+1)k^k}{(k+1)!}[/math](writing the last [math]1[/math] as [math]\dfrac{(k+1)k(k-1)(k-2)\cdots 1}{(k+1)!}[/math], and replacing every [math](k-1),(k-2),(k-3),[/math] etc. in the entire expression by [math]k[/math][math][/math])[math]=k^{k+1}+(k+1)k^k+\dfrac{k+1}{2!}k^k+\dfrac{k+1}{3!}k^k+\dfrac{k+1}{4!}k^k+\cdots+\dfrac{(k+1)k^k}{(k+1)!}[/math][math]<k^k\left((k+1)+(k+1)+\frac{k+1}{2!}+\frac{k+1}{3!}+\cdots+\frac{k+1}{(k+1)!}\right)[/math][math]=(k^k)(k+1)\left(1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{(k+1)!}\right)[/math][math]<(k^k)(k+1)e[/math]since last bracket is a partial sum of [math]e=1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots[/math])[math]<3(k+1)(k^k)[/math]since [math]e<3[/math][math]<3(k+1)(3^k\,k!)[/math]by the inductive hypothesis[math]=3^{k+1}(k+1)!.[/math]The induction is complete.