3

show that:

$$\lim_{n\to\infty}n\left[\left(\dfrac{1}{\pi}\left(\sin{\left(\dfrac{\pi}{\sqrt{n^2+1}}\right)}+ \sin{\left(\dfrac{\pi}{\sqrt{n^2+2}}\right)}+\cdots+\sin{\left(\dfrac{\pi}{\sqrt{n^2+n}}\right)} \right)\right)^n-\dfrac{1}{\sqrt[4]{e}}\right]=-\dfrac{1}{\sqrt[4]{e}}\left(\dfrac{31}{96}+\dfrac{\pi^2}{6}\right)$$

I only solve this :$$\lim_{n\to\infty}\left(\dfrac{1}{\pi}\sum_{i=1}^{n}\sin{\left(\dfrac{\pi}{\sqrt{n^2+i}}\right)}\right)^n=\dfrac{1}{\sqrt[4]{e}}\tag{1}$$ use $$\sin{x}\approx x,\Longrightarrow \dfrac{1}{\pi}\sin{(\dfrac{\pi}{\sqrt{n^2+i}})}=\dfrac{1}{\sqrt{n^2+i}}+o(\dfrac{1}{\sqrt{n^2+i}})\to \dfrac{1}{n}\left(1+\dfrac{i}{n^2}\right)^{-\frac{1}{2}},n\to\infty$$ and note $$(1+x)^{-1/2}=1-\dfrac{1}{2}x+o(x)$$ so $$\lim_{n\to\infty}\left(\dfrac{1}{\pi}\sum_{i=1}^{n}\sin{\left(\dfrac{\pi}{\sqrt{n^2+i}}\right)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1}{n}\sum_{i=1}^{n}\left(1-\dfrac{i}{2n^2}+o(i/n^2)\right)\right)^n=e^{-\frac{1}{4}}$$

But I can't solve my problem,Thank you

math110
  • 93,304

2 Answers2

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The binomial theorem says $$ \begin{align} \left(1+\frac{k}{n^2}\right)^{-1/2} &=1-\frac12\frac{k}{n^2}+\frac38\frac{k^2}{n^4}+O\left(\frac{k^3}{n^6}\right)\\ \left(1+\frac{k}{n^2}\right)^{-3/2} &=1+O\left(\frac{k}{n^2}\right) \end{align} $$ We can use the Taylor series for $\sin(x)$ to get $$ \begin{align} \sin\left(\frac{\pi}{\sqrt{n^2+k}}\right) &=\frac\pi{n}\left(1+\frac{k}{n^2}\right)^{-1/2} -\frac16\frac{\pi^3}{n^3}\left(1+\frac{k}{n^2}\right)^{-3/2} +O\left(\frac1{n^5}\right)\\ &=\pi\frac1n-\frac\pi2\frac{k}{n^3}+\frac{3\pi}8\frac{k^2}{n^5}+O\left(\frac{k^3}{n^7}\right)\\ &-\frac{\pi^3}6\frac1{n^3}+O\left(\frac{k}{n^5}\right)\\ &+O\left(\frac1{n^5}\right) \end{align} $$ Sum to get $$ \begin{align} \sum_{k=1}^n\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right) &=\pi\frac{n}n-\frac\pi4\frac{n^2+n}{n^3}+\frac\pi{16}\frac{2n^3+3n^2+n}{n^5}+O\left(\frac1{n^3}\right)\\ &-\frac{\pi^3}6\frac{n}{n^3}+O\left(\frac1{n^3}\right)\\ &+O\left(\frac1{n^4}\right)\\ &=\pi-\frac\pi4\frac1n-\left(\frac\pi8+\frac{\pi^3}6\right)\frac1{n^2}+O\left(\frac1{n^3}\right) \end{align} $$ Divide by $\pi$ and use the power series for $\log(1+x)$ and $e^x$ $$ \begin{align} \log\left(\frac1\pi\sum_{k=1}^n\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right) &=-\frac14\frac1n-\left(\frac18+\frac{\pi^2}6\right)\frac1{n^2}+O\left(\frac1{n^3}\right)\\ &-\frac1{32}\frac1{n^2}+O\left(\frac1{n^3}\right)\\ &+O\left(\frac1{n^3}\right)\\ &=-\frac14\frac1n-\left(\frac5{32}+\frac{\pi^2}6\right)\frac1{n^2}+O\left(\frac1{n^3}\right)\\ n\log\left(\frac1\pi\sum_{k=1}^n\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right) &=-\frac14-\left(\frac5{32}+\frac{\pi^2}6\right)\frac1n+O\left(\frac1{n^2}\right)\\ \left(\frac1\pi\sum_{k=1}^n\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right)^n &=e^{-1/4}\left(1-\left(\frac5{32}+\frac{\pi^2}6\right)\frac1n+O\left(\frac1{n^2}\right)\right)\\ n\left[\left(\frac1\pi\sum_{k=1}^n\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right)^n-e^{-1/4}\right] &=e^{-1/4}\left(-\left(\frac5{32}+\frac{\pi^2}6\right)+O\left(\frac1n\right)\right)\\ &\to-\left(\frac5{32}+\frac{\pi^2}6\right)e^{-1/4} \end{align} $$

My answer agrees with Paramanand Singh's. As a check, with $n=1000000$ $$ \begin{align} n\left[\left(\frac1\pi\sum_{k=1}^n\sin\left(\frac{\pi}{\sqrt{n^2+k}}\right)\right)^n-e^{-1/4}\right]&=-1.4027616010762190386\\ -\left(\frac5{32}+\frac{\pi^2}6\right)e^{-1/4}&=-1.4027635617171364019\\ -\left(\frac{31}{96}+\frac{\pi^2}6\right)e^{-1/4}&=-1.5325636922290372132 \end{align} $$ It looks as if the question is in error.

robjohn
  • 345,667
  • Thanks robjohn for helping me out. After much effort I couldn't find an issue in my calculation. Your calculation is much more detailed and also indicates the error terms properly. +1 – Paramanand Singh Nov 08 '14 at 12:27
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You seem to be on the right track and I guess we can try to get more reasonable estimates using further terms of the expansion. First we need to use $\sin x \approx x - \dfrac{x^{3}}{6}$ to get $$\begin{aligned}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right) &\approx \frac{1}{n}\left(1 + \frac{i}{n^{2}}\right)^{-1/2} - \frac{\pi^{2}}{6n^{3}}\left(1 + \frac{i}{n^{2}}\right)^{-3/2}\\ &\approx \frac{1}{n}\left(1 - \frac{i}{2n^{2}} + \frac{3i^{2}}{8n^{4}}\right) - \frac{\pi^{2}}{6n^{3}}\left(1 - \frac{3i}{2n^{2}} + \frac{15i^{2}}{8n^{4}}\right)\\ &\approx \frac{1}{n}\left(1 - \frac{3i + \pi^{2}}{6n^{2}} + \frac{9i^{2} + 6i\pi^{2}}{24n^{4}}\right)\end{aligned}$$ and then summing from $i = 1$ to $i = n$ we get $$\begin{aligned}\sum_{i = 1}^{n}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right) &\approx \frac{1}{n}\left(n - \dfrac{\dfrac{3n(n + 1)}{2} + n\pi^{2}}{6n^{2}} + \dfrac{\dfrac{3n(n + 1)(2n + 1)}{2} + 3n(n + 1)\pi^{2}}{24n^{4}}\right)\\ &= \left(1 - \frac{3(n + 1) + 2\pi^{2}}{12n^{2}} + \frac{(n + 1)(2n + 1) + 2(n + 1)\pi^{2}}{16n^{4}}\right)\\ &\approx \left(1 - \frac{1}{4n} - \frac{2\pi^{2} + 3}{12n^{2}} + \frac{1}{8n^{2}}\right)\\ &= \left(1 - \frac{1}{4n} - \frac{4\pi^{2} + 3}{24n^{2}}\right)\\\end{aligned}$$ and therefore using logs we get $$\begin{aligned}\left(\sum_{i = 1}^{n}\frac{1}{\pi}\sin\left(\frac{\pi}{\sqrt{n^{2} + i}}\right)\right)^{n} &\approx \exp\left(n\log\left(1 - \frac{1}{4n} - \frac{4\pi^{2} + 3}{24n^{2}}\right)\right)\\ &\approx\exp\left(-n\left(\frac{1}{4n} + \frac{\pi^{2}}{6n^{2}} + \frac{5}{32n^{2}}\right)\right)\\ &= \exp\left(-\frac{1}{4} - \frac{\pi^{2}}{6n} - \frac{5}{32n}\right)\\ \end{aligned}$$ We can now see that the desired limit is equal $$\begin{aligned}L &= \lim_{n \to\infty}n\left(\exp\left(-\frac{1}{4} - \frac{\pi^{2}}{6n} - \frac{5}{32n}\right) - \exp(-1/4)\right)\\ &= e^{-1/4}\lim_{n \to\infty}n\left(\exp\left(- \frac{\pi^{2}}{6n} - \frac{5}{32n}\right) - 1\right) = -\frac{1}{\sqrt[4]{e}}\left(\frac{\pi^{2}}{6} + \frac{5}{32}\right)\end{aligned}$$ There might be some calculation mistake in above or there is some typo in the question.

Paramanand Singh
  • 87,309
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    Where are all the necessary controls on the terms you neglect? – Did Nov 07 '14 at 15:59
  • @Did: from the form of the limit expression it is clear that we need to consider the terms up to the order of $1/n^{2}$ and the argument can be easily justified by writing $o(1/n^{2})$ for the terms neglected. When raising to power $n$ this leads to terms up to order of $1/n$ and thats what we need in the last step. – Paramanand Singh Nov 07 '14 at 16:23
  • No. For example, in the first expansion of the sines, you are keeping terms of order $1/n^5$ hence using $\sin(x)\approx x-x^3/6$ with $x$ of order $1/n$ is not sufficient. – Did Nov 07 '14 at 16:46
  • @Did: even if I use $\sin x=x-(x^3/6)+(x^5/120)$ I will get a term like $\pi^{4}/120n^{5}$ and it will not contribute to any term of order $1/n$ in final expression dealing with $\exp$. In fact the term $i \pi^{2}/4n^{5}$ which I have considered in my calculation does not contribute anything in the final expression for $\exp$. – Paramanand Singh Nov 07 '14 at 18:52
  • Then why mention it? The trouble comes from a non rigorous use of the symbol $\approx$, which leaves open the door for all sorts of forgotten terms. – Did Nov 07 '14 at 22:00
  • @Did: I think your main concern is regarding the use of $\approx$. It would have been better had I written $= \cdots + o(1/n^{k})$ for some suitable $k$. By the way I am unable to find any missing terms which may have impact on the final result. – Paramanand Singh Nov 08 '14 at 05:05
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    We cannot know--unless one keeps track rigorously of the terms neglected, which is not done at present. This is my point since the very first comment. – Did Nov 08 '14 at 08:33
  • Let us continue this discussion in chat. – Paramanand Singh Nov 08 '14 at 08:43
  • @ParamanandSingh: Luckily, no terms contributing significantly to the result were missed. However, Did is correct: to be sure, one must keep track of the error terms. – robjohn Nov 08 '14 at 10:33