1

For each $n\in\mathbb{N}$ consider $$W_n=\int_0^{\pi/2}\cos^n{t} dt$$ We know the that $$\lim_{n\to\infty}\dfrac{W_n}{\sqrt{\dfrac{\pi}{2n}}}=1$$ Find the radius of convergence and the sum of the series: $$\sum_{n=0}^{\infty}W_nx^n$$ What I have obtained is that $W_n=\dfrac{n-1}{n}\dfrac{n-3}{n-2}...\dfrac{2}{3}$

PaulDirac
  • 879
  • 1
    Since $W_n \sim \sqrt{\frac{\pi}{2n}}$ the convergence radius of the series is the same as that of $\sum \frac{x^n}{\sqrt{n}}$. – Winther Feb 01 '17 at 15:47
  • For evaluating the sum one approach that often works is by starting with the recursion formula that lead you to the last result. Then multiply it by $x^n$ and get it on the form $f(n)W_nx^n = g(n) W_{n-2}x^n$ for some (linear I think) polynomials $f,g$. Then sum over all $n$ and relate the different terms to $F(x) = \sum W_n x^n$ and it's derivatives. This will give you a ODE for $F(x)$ which hopefully can be solved. – Winther Feb 01 '17 at 15:58
  • 1
    ...and a simpler way is to use the integral in the series: $$F(x) = \sum_{n=0}^\infty \int_0^{\pi/2}(\cos(t) x)^n{\rm d}t = \int_0^{\pi/2} \frac{1}{1-x\cos(t)}{\rm d}t$$ where we interchanged the summation and integration and summed a geometrical series. – Winther Feb 01 '17 at 16:13
  • Very ingenious solutions. But why are you able to commute summation with the integral? – PaulDirac Feb 01 '17 at 16:17
  • See this question for some useful criteria. In this case we justify it from the uniform convergence of $\sum f_n(t)$ with $f_n(t) = (\cos(t)x)^n$ for $|x|<1$. – Winther Feb 01 '17 at 16:25
  • I see. Thank you very much! – PaulDirac Feb 01 '17 at 17:08

1 Answers1

3

As suggested by Winther, $$\begin{eqnarray*} \sum_{n\geq 0} x^n\int_{0}^{\pi/2}\cos^n(t)\,dt &=& \int_{0}^{\pi/2}\frac{dt}{1-x\cos(t)}\\&=&2\int_{0}^{\pi/4}\frac{dt}{(1+x)-2x\cos^2(t)}\\&=&2\int_{0}^{1}\frac{dt}{(1+x)(1+t^2)-2x}\\&=&\frac{2}{\sqrt{1-x^2}}\,\arctan\sqrt{\frac{1+x}{1-x}}\\&=&\frac{\pi+2\arcsin(x)}{2\sqrt{1-x^2}}=\frac{\pi-\arccos(x)}{\sqrt{1-x^2}}\end{eqnarray*}$$ whose radius of convergence is clearly $1$, due to the singularity at $x= 1$. On the other hand we do not need to explicitly locate such singularities to be able to state the same, since: $$ \int_{0}^{\pi/2}\cos^n(t)\,dt \sim\sqrt{\frac{\pi}{2n}} $$ ensures $\rho=1$ through the definition of the radius of convergence as a $\limsup$.

Jack D'Aurizio
  • 353,855