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Let $$f(x)=\sum_{n=1}^{\infty}\dfrac{\cos{nx}}{\sqrt{n^3+n}}$$ and $F(x)=\int_{0}^{x}f(t)\,\mathrm dt,F(0)=0$.


  1. Show that: $$\dfrac{\sqrt{2}}{2}-\dfrac{1}{15}<F\left(\dfrac{\pi}{2}\right)<\dfrac{\sqrt{2}}{2}$$

  2. Find the value $F\left(\dfrac{\pi}{2}\right)$.

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  • $\begingroup$ about (2) is my add it,Thank you @julien $\endgroup$ – math110 May 28 '13 at 14:15
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    $\begingroup$ @ThomasAndrews In the future, instead of making one-character edits, please either take the time to improve the whole post, or just leave it alone for someone else. $\endgroup$ – Lord_Farin May 28 '13 at 14:21
  • $\begingroup$ Thank you every much@ThomasAndrews $\endgroup$ – math110 May 28 '13 at 14:22
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    $\begingroup$ I can reduce $F(\pi/2)$ to $\sum_{n=1}^{\infty}\dfrac{(-1)^n}{\sqrt{8n^3+2n}}$ $\endgroup$ – ratchet freak May 28 '13 at 14:26
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    $\begingroup$ @Lord_Farin Since you care about edits, what is the point of highlighting the content of the whole post? $\endgroup$ – Julien May 28 '13 at 15:09
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Since the series $f$ converges normally on $\mathbb{R}$, we can integrate term-by-term. This gives $$ F\left(\frac{\pi}{2}\right)=\sum_{n\geq 1}\frac{1}{\sqrt{n^3+n}}\int_0^{\pi/2}\cos (nx)dx=\sum_{k\geq 0}\frac{(-1)^k}{\sqrt{2}(2k+1)^\frac{3}{2}(2k^2+2k+1)^\frac{1}{2}}. $$ The series converges absolutely to $S$. But this is an alternating series which satisfies Leibniz criterion (the relevant function is indeed decreasing on $(0,+\infty)$). So the sequence of partial sums $S_n$ alternates about $S$ and satisfies $$ S_{2n+1}< S< S_{2n} \qquad \forall n\geq 0. $$ In particular, we get $$ \frac{\sqrt{2}}{2}-\frac{1}{15}<\frac{\sqrt{2}}{2}-\frac{1}{3\sqrt{5}\sqrt{6}}=S_1<S<S_0=\frac{\sqrt{2}}{2}. $$ This answers 1.

I don't know if there is a closed form for $S$ and I am not alone.

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    $\begingroup$ Thank you,@julien, I think maybe this problem (a) has other nice methos. $\endgroup$ – math110 May 28 '13 at 14:42
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    $\begingroup$ Nice answer (+1). $\endgroup$ – Mhenni Benghorbal May 28 '13 at 16:37
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    $\begingroup$ My answer adds little to yours (+1), but I also tried to look up the answer on the ISC. $\endgroup$ – robjohn May 28 '13 at 17:36
  • $\begingroup$ @math110: which problem has other nice methods, and what might those methods be? $\endgroup$ – robjohn May 28 '13 at 17:37
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    $\begingroup$ @robjohn Thanks. And I did not know the ISC...! $\endgroup$ – Julien May 28 '13 at 17:58
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I was delayed in posting this, so it is little more than a comment to julien's answer.

Since the series $\sum\limits_{n=1}^\infty\frac1{\sqrt{n^3+n}}$ is convergent, we can integrate term by term $$ F(x)=\sum_{n=1}^\infty\frac{\sin(nx)}{n\sqrt{n^3+n}} $$ $\sin((2k)\pi/2)=0$ and $\sin((2k+1)\pi/2)=(-1)^k$; therefore, $$\begin{align} F(\pi/2) &=\frac1{\sqrt2}\sum_{k=0}^\infty\frac{(-1)^k}{\sqrt{(2k+1)^3(2k^2+2k+1)}}\\ &=\frac1{\sqrt2}-\frac1{\sqrt{270}}+\dots \end{align} $$ By the alternating series test, the final sum is between $\frac1{\sqrt2}$ and $\frac1{\sqrt2}-\frac1{\sqrt{270}}$. That is, $$ \frac1{\sqrt2}-\frac1{15}\lt\frac1{\sqrt2}-\frac1{\sqrt{270}}\lt F(\pi/2)\lt\frac1{\sqrt2} $$

This evaluates to $0.6587329279592957$, but the ISC does not find a closed form.

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