1. How to prove:

$$\int_0^1\left(1+\sin \dfrac{\pi x}{2} \right)^ndx>\dfrac{2^{n+1}-1}{n+1},\quad n\in \mathbb{N^+}$$

  1. and find the limit below:

$$\lim\limits_{n\rightarrow+\infty}\left(\int_0^1\left(1+\sin \dfrac{\pi x}{2} \right)^ndx\right)^{\frac{1}{n}}$$

Both are from the undergraduate admission exam of Fudan university.


You have $\sin(x)\geq \frac{2x}{\pi}$ on $[0,\pi/2]$ (Use the fact that the function $\sin(x)$ is concave on $I$, and that the chord from $A=(0,0)$ to $B=(\pi/2,1)$ is below the curve $y=\sin(x)$). Now on $[0,1]$, you get that $\sin(\frac{\pi x}{2})\geq x$, and $(1+\sin(\frac{\pi x}{2}))^n\geq (1+x)^n$, hence integrating from $0$ to $1$, you get the inequality with $\geq$. If the two terms are equals, then if $g(x)=(1+\sin(\frac{\pi x}{2}))^n -(1+x)^n$, we get that $\int_0^1 g(x)dx=0$, and as $g$ is continuous and $\geq 0$, $g(x)=0$ for all $x$, this is not true.

For the second question, note that if $u_n=\int_0^1(1+\sin(\frac{\pi x}{2}))^n dx$, then it is obvious that $u_n\leq 2^n$, and with the inequality in 1), it is easy to finish.

  • 1
    $\begingroup$ Well done! All too easy. (+1) $\endgroup$ – Mark Viola Jul 11 '17 at 15:56

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