$$u_n = \int_{0}^{\frac{\pi}{2}} \sin^n(x)dx$$ for $ n \in \mathbb{N}^*$.
- Prove that $(u_n)$ is convergent toward $0$.
- Prove that the series with the general term $(-1)^n u_n$ converges.
- Prove that $$\sum_{n=0}^{\infty} (-1)^n u_n = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx$$
Compute $$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx$$
Hint: You can start by proving $$\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin(x)}dx = \int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \cos(x)}dx$$
Prove that the series with the general term $u_n$ diverges.
Hint: You can start by proving $$u_n \geq \frac{1}{n + 1}$$
- For $ 0 \leq x \leq \frac{\pi}{2}$ we have: $0 \leq u_n \leq \frac{\pi}{2}$ wich I can't derive the convergence from it. How can I prove it is convergent toward 0?
I am stuck with question 4. and 5, I could not see how to prove the hints.
Best Answer
Using integration by parts it is easy to prove that $u_n=((n-1)u_{n-2})/n$. Using this relation it follows that $u_n$ is decreasing and clearly it is non-negative so that $u_n$ is convergent. Using this recurrence relation we will prove that sequence $a_{n} =u_{2n}$ tends to $0$. We have $$\frac{a_{n}} {a_{n-1}}=\frac{2n-1}{2n}$$ and hence on taking logs we have $$\log a_n-\log a_{n-1}=\log\left(1-\frac{1}{2n}\right)<-\frac{1}{2n}$$ And on adding such equations we get $$\log a_n-\log a_0<-\frac{1}{2}\sum_{i=1}^{n}\frac{1}{i}$$ Since the expression on RHS diverges to $-\infty$ it follows that $\log a_n\to-\infty $ and therefore $a_n\to 0$. Since $a_n$ is a subsequence of a convergent sequence $u_n$ it follows that $u_n\to 0$.
From the recurrence relation for $u_n$ it follows that it is decreasing and since $u_n\to 0$ by Leibniz test the alternating series $\sum (-1)^nu_n$ converges.
Next note that $$\sum_{i=0}^{n}(-1)^iu_i=\int_{0}^{\pi/2}\sum_{i=0}^{n}(-1)^{i}\sin^ix\,dx$$ The integral on right side can be written as $$\int_{0}^{\pi/2}\frac{dx}{1+\sin x} +(-1)^{n+1}R_n$$ where $$0\leq R_n=\int_{0}^{\pi/2}\frac{\sin^{n+1}x}{1+\sin x} \, dx\leq \int_{0}^{\pi/2}\sin^{n+1}x\,dx=u_{n+1}$$ and therefore by squeeze theorem $R_n\to 0 $. And thus $(-1)^{n+1}R_n\to 0$. It follows that $$\sum_{n=0}^{\infty} (-1)^nu_n=\int_{0}^{\pi/2}\frac{dx}{1+\sin x} $$ The integral above can be easily evaluated by using standard substitution $t=\tan (x/2)$ and the integral value is $1$.
The final problem is the divergence of $\sum u_n$ which is easily handled by noting that $$u_n=\int_{0}^{1}\frac{t^n}{\sqrt{1-t^2}}\,dt\geq \int_{0}^{1}t^n\,dt=\frac{1}{n+1}$$ (also mentioned in comments to question by Kavi Rama Murthy).