$$\displaystyle{\ln{2}-\frac{1}{\sqrt{n\pi}}<\sum_{k=1}^{n}{\frac{(2k-1)!!}{2k(2k)!!}}<\ln{2}}.$$ I thought of using Wallis inequality,I get the following result $$\frac{1}{\sqrt{(k+1)\pi}}<\frac{(2k-1)!!}{(2k)!!}<\frac{1}{\sqrt{k\pi}}$$ $$\sum_{k=1}^{n}\frac{1}{2k\sqrt{(k+1)\pi}}<\sum_{k=1}^{n}{\frac{(2k-1)!!}{2k(2k)!!}}<\sum_{k=1}^{n}\frac{1}{2k\sqrt{k\pi}}<\sum_{k=1}^{n}\frac{1}{n+k}$$ Now I don't know how to get the estimate on the left side of the inequality. I want to confirm whether my right side is correct. Thanks!
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1$\displaystyle\sum_{k=1}^{n}\frac{1}{2k\sqrt{k\pi}}<\sum_{k=1}^{n}\frac{1}{n+k}$ can't be correct: the limit as $n\to\infty$ of the LHS is $\displaystyle\frac{\zeta(3/2)}{2\sqrt{\pi}}\approx0.73693748002264514384792331234796508152$, that's bigger than the limit of the RHS, $\ln2$. – Jan 06 '21 at 08:20
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@Professor Vector Sorry, Do you mean the inequality on the right doesn't hold？ – Hilbert1994 Jan 06 '21 at 08:28
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The rightmost inequality doesn't hold, that's what I wrote, indeed. Your idea is good, but you should use it to estimate the remainder of the series, $\displaystyle\sum_{k=n+1}^\infty{\frac{(2k-1)!!}{2k(2k)!!}}$. – Jan 06 '21 at 08:33
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1@Professor Vector Thank you. According to your tips, I got it$$\displaystyle\sum_{k=n+1}^\infty{\frac{(2k-1)!!}{2k(2k)!!}}<\frac{1}{\sqrt{n\pi}}.$$In fact, the problem was originally for calculation$$\displaystyle\sum_{k=1}^\infty{\frac{(2k-1)!!}{2k(2k)!!}}$$But I don't want to use term by term integration and term by term derivation. I want to try to use inequality to get estimates. – Hilbert1994 Jan 06 '21 at 08:45
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@yf1994 However, you still have to prove the limit of the series is $\ln 2$. – Jean-Claude Arbaut Jan 06 '21 at 08:59
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@Jean-Claude Arbaut I can use the power series to find out that the sum of the series is $\ln2$, but I can't think of how to use the inequality to estimate – Hilbert1994 Jan 06 '21 at 09:04
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Once you have the inequality in your previous comment, simply add $\sum_{k=1}^n (2k-1)!!/(2k(2k)!!)$ to both sides. You get $\ln2<\sum_{k=1}^n (2k-1)!!/(2k(2k)!!)+1/\sqrt{n\pi}$, that is equivalent to the left inequality in your question. The right inequality is immediate given the sum of the series in $\ln2$. – Jean-Claude Arbaut Jan 06 '21 at 09:14
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@Jean-Claude Arbaut Thank you. It's my fault.So there's no way to estimate the sum of series by inequality? – Hilbert1994 Jan 06 '21 at 09:20
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Well, comparing a series to an integral is a classic approach, and Wallis' inequality follows from an integral anyway. What's the problem? – Jean-Claude Arbaut Jan 06 '21 at 09:22
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@Jean-Claude Arbaut Oh sorry, my mistake – Hilbert1994 Jan 06 '21 at 09:30
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@Jean-Claude Could you please help me with this problem? Thank you very much https://math.stackexchange.com/questions/3978738/how-to-prove-this-problem-about-uniform-continuity – Hilbert1994 Jan 10 '21 at 04:48
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@Jean-Claude Could you please help me with this problem? Thank you very much https://math.stackexchange.com/questions/3981252/how-to-understand-the-following-derivation-result – Hilbert1994 Jan 12 '21 at 00:20