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Please find the limit:$$\mathop {\lim }\limits_{n \to \infty } n\left[ {{{\left( {\frac{1}{\pi }\left( {\sin \left( {\frac{\pi }{{\sqrt {{n^2} + 1} }}} \right) + \sin \left( {\frac{\pi }{{\sqrt {{n^2} + 2} }}} \right) + \cdots+ \sin \left( {\frac{\pi }{{\sqrt {{n^2} + n} }}} \right)} \right)} \right)}^n} - \frac{1}{{\sqrt[4]{e}}}} \right].$$

Eufisky
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    I see a lot of problems like this one on stack exchange (the integrals, limits, or sums of giant expressions with many functions involved), but I don't understand what makes them seem like interesting or fun problems to solve - they look like a lot of messy computation to me ... – Zubin Mukerjee Nov 08 '14 at 12:50
  • @ZubinMukerjee Maybe,but it's interesting for me,and so far,I haven't found a available method to solve it. – Eufisky Nov 08 '14 at 12:56
  • $\sin x = x - \frac{x^3}{6} + O(x^5)$, and some - probably somewhat messy - computation. You want the stuff that is raised to the $n$-th power as $1 + \frac{a}{n} + \frac{b}{n^2} + O(n^{-3})$. – Daniel Fischer Nov 08 '14 at 12:58
  • @ZubinMukerjee Could it be that they are given by teachers as exercises, with the aim to not let their students get away with mathematica or wolfram alpha? – Student tea Nov 08 '14 at 13:17

2 Answers2

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Step 1. Since for small $x$ we have $\frac{1}{\sqrt{1+x}}=1-\frac{x}{2}+\frac{3x^2}{8}+O(x^3),$ $$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{1+\frac{k}{n^2}}}=\frac{1}{n}\sum_{k=1}^{n}\left(1-\frac{k}{2n^2}+\frac{3k^2}{8n^4}\right)+O\left(\frac{1}{n^3}\right)$$ hence: $$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=1-\frac{1}{4n}-\frac{1}{8n^2}+O\left(\frac{1}{n^3}\right).$$ Step 2. $$\sum_{k=1}^{n}\frac{1}{\left(\sqrt{n^2+k}\right)^3}=\frac{1}{n^2}+O\left(\frac{1}{n^3}\right).$$ Step 3. Since for small $x$ we have $\sin x = x-\frac{x^3}{6}+O(x^5)$, $$\sum_{k=1}^{n}\sin\frac{\pi}{\sqrt{n^2+k}}=\pi-\frac{\pi}{4n}-\frac{\pi}{8n^2}-\frac{\pi^3}{6n^2}+O\left(\frac{1}{n^3}\right).$$ Step 4. Since for small $x$ we have $\log(1+x)=x-\frac{x^2}{2}+O(x^3)$, $$ n\log\left(\frac{1}{\pi}\sum_{k=1}^{n}\sin\frac{\pi}{\sqrt{n^2+k}}\right)=-\frac{1}{4}-\left(\frac{5}{32}+\frac{\pi^2}{6}\right)\frac{1}{n}+O\left(\frac{1}{n^2}\right).$$ Step 5. Exponentiating, it follows that the value of the limit is:

$$ L = -\frac{15+16\,\pi^2}{96\, e^{1/4}}.$$

Jack D'Aurizio
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Since $n$ is supposed to be large, then the argument of the sine is small. So, start with $$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^6\right)$$ Now, replace $x$ by $\frac{\pi }{\sqrt{k+n^2}}$ and then $$\frac{1 }{\sqrt{k+n^2}}=\frac{1}{n}-\frac{k}{2 n^3}+\frac{3 k^2}{8 n^5}+O\left(\left(\frac{1}{n}\right)^6\right)$$ So, $$\sin \left(\frac{\pi }{\sqrt{k+n^2}}\right)\approx\frac{\pi }{n}-\frac{\frac{\pi k}{2}+\frac{\pi ^3}{6}}{n^3}$$ Now, sum over $k$ from $1$ to $n$ and the sum of sines is $$\pi-\frac{\pi }{4 n}-\frac{\pi \left(3+2 \pi ^2\right)}{12 n^2}\approx \pi-\frac{\pi }{4 n}$$ So, what is inside brackets write $$\left(1-\frac{1}{4 n}\right)^n-\frac{1}{\sqrt[4]{e}}$$

I am sure that you recognize the first term and that you can take from here.

Claude Leibovici
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