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It's given that $$I_n=\int_0^{\frac{\pi}{2}}\frac{\sin^2nx}{\sin x}dx,$$and I need to prove that $$I_n\leq1+\ln\sqrt{2n-1}.$$ I have got that $$I_n-I_{n-1}=\int_0^{\frac{\pi}{2}}\frac{\sin^2nx}{\sin x}dx-\int_0^{\frac{\pi}{2}}\frac{\sin^2(n-1)x}{\sin x}dx\\=\int_0^{\frac{\pi}{2}}\frac{[\sin nx-\sin(n-1)x][\sin nx+\sin(n-1)x]}{\sin x}dx\\=\int_0^{\frac{\pi}{2}}\frac{2\cos\frac{(2n-1)x}{2}\sin\frac{x}{2}\cdot2\sin\frac{(2n-1)x}{2}\cos\frac{x}{2}}{\sin x}dx\\=\int_0^{\frac{\pi}{2}}\frac{\sin[(2n-1)x]\sin x}{\sin x}dx\\=\int_0^{\frac{\pi}{2}}\sin[(2n-1)x]dx\\=\frac{1}{2n-1}.$$ Then it's easy to know that $$I_n=\sum_{r=1}^{n}\frac{1}{2r-1}.$$ That's all the work I've done. What should I do next?

CESTU
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    You're basically done. Define the harmonic numbers $H_n = \sum_{i\le n} 1/i.$ Then $I_n = H_{2n-1} - 0.5 H_{n-1}$. Standard bounds on the Harmonic numbers, e.g. this, show that $ 0.5 + \log n \le H_n \le 1 + \log n, $ from which the bound you want to show follows (you might need to use $\log 2 < 1$ somewhere). You can probably also directly show the bound using techniques in the answers to the question I linked. – stochasticboy321 Mar 17 '24 at 14:29

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By induction: For $n=1,2$, $I_1=1,I_2=\frac{4}{3}<1+\ln\sqrt{3}$ is easy to check. Suppose $I_n<1+\ln\sqrt{2n-1}$, then $$I_{n+1}=I_n+\frac1{2n+1}<1+\ln\sqrt{2n-1}+\frac1{2n+1}.$$ We need to prove $$1+\ln\sqrt{2n-1}+\frac1{2n+1}<1+\ln\sqrt{2n+1},$$ which is equivalent to $$\frac1{2n+1}<\frac{1}{2}\ln\frac{2n+1}{2n-1}.$$ This follows from the inequality: $$\frac{x}{1+x}<\ln(1+x),\quad \forall x>0,$$ just take $x=\frac{2}{2n-1}$.

Riemann
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