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Prove that $$\lim_{n\to\infty}(g(n+1)-g(n))$$ exists when

$$g(n)=16^n\left(1+2\sum_{j=1}^n (-1)^j \cos^{2n}\left(\frac{j \pi}{2n+1}\right)\right)^2$$ I can't see how to do this. What' the first step?

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    $\begingroup$ First step would be to find the sum using geometric series and Euler formula connecting exponential and circular functions. $\endgroup$ – Paramanand Singh Aug 13 '17 at 5:40
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    $\begingroup$ @ParamanandSingh Are you saying the sum has a closed form? $\endgroup$ – MeMyselfI Aug 15 '17 at 8:18
  • $\begingroup$ I have not performed the calculations, but I suspect that there should be a closed form. $\endgroup$ – Paramanand Singh Aug 15 '17 at 8:21
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1. Preliminary

We introduce some facts and observations to be used throughout the computation. This may as well help those who are already familiar with such computations and want to skip the details.

1.1. Dirichlet kernel. By a simple application of the geometric sum formula, we have

$$ \sum_{k=-n}^{n} e^{\mathrm{i}kx} = \frac{\sin\left(\frac{2n+1}{2}x\right)}{\sin\left(\frac{1}{2}x\right)}. \tag{1.1} $$

1.2. Beta function identity. For $\operatorname{Re}(z) > 0$ and $\operatorname{Re}(w) > 0$ we have

$$ \int_{0}^{1} x^{z-1}(1-x)^{w-1} \, dx = \frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}, \tag{1.2} $$

where $\Gamma$ is the gamma function.

1.3. An integral identity. For $|\Re(s)| < \frac{1}{2}$, we have

$$ \sec(\pi s) = \frac{2}{\pi} \int_{0}^{\infty} \frac{u^{2s}}{1+u^2} \, du = \frac{2}{\pi} \int_{0}^{1} \frac{u^{2s} + u^{-2s}}{1+u^2} \, du. \tag{1.3}$$

This can be obtained directly by adopting some contour integration technique. Alternatively, this can be derived by plugging the substitution $x = \frac{1}{1+u^2}$ to $\text{(1.2)}$ with $z = s = 1-w$ and applying the Euler's reflection formula.


2. Solution

2.1. Derivation of the key formula. Notice that we can write $g(n) = S_n^2$, where $S_n$ is given by

$$S_n = 4^n\left( 1 + 2\sum_{j=1}^{n}(-1)^j \cos^{2n}\left(\frac{j\pi}{2n+1}\right)\right).$$

We perform some algebraic manipulation on $S_n$ by utilizing the identity $2\cos x = e^{\mathrm{i}x} + e^{-\mathrm{i}x}$ and the binomial theorem.

\begin{align*} S_n &= 4^n \sum_{j=-n}^{n} (-1)^j \cos^{2n}\left(\frac{j\pi}{2n+1}\right) \\ &= \sum_{j=-n}^{n} (-1)^j \left( 2\cos\left(\frac{j\pi}{2n+1}\right) \right)^{2n}\\ &= \sum_{j=-n}^{n} (-1)^j \sum_{l=0}^{2n}\binom{2n}{l} \exp\left(\mathrm{i}\frac{(2l-2n)j\pi}{2n+1} \right). \end{align*}

Interchanging the summations and applying $\text{(1.1)}$ and $\text{(1.3)}$ successively, this simplifies to

\begin{align*} S_n &\stackrel{\text{(1.1)}}{=} \sum_{l=0}^{2n}\binom{2n}{l} (-1)^{l} \sec\left(\frac{(l-n)\pi}{2n+1}\right) \\ &\stackrel{\text{(1.3)}}{=} \frac{2}{\pi} \sum_{l=0}^{2n}\binom{2n}{l} (-1)^{l} \int_{0}^{1} \frac{u^{2(l-n)/(2n+1)} + u^{-2(l-n)/(2n+1)}}{1+u^2} \, du. \end{align*}

Applying the binomial theorem and plugging the substitution $u = v^{2n+1}$, we obtain the following integral representation which is a key step to our goal.

\begin{align*} S_n = \frac{4}{\pi} \int_{0}^{1} \frac{\left(1-u^{2/(2n+1)} \right)^{2n}}{u^{2n/(2n+1)} (1+u^2)} \, du = \bbox[border:2px dashed green,6px]{ \frac{4(2n+1)}{\pi} \int_{0}^{1} \frac{\left(1-v^2 \right)^{2n}}{1+v^{4n+2}} \, dv }. \tag{2.1} \end{align*}

2.2. Evaluation of the limit. We estimate our integral representation $\text{(2.1)}$ as follows:

\begin{align*} \int_{0}^{1} \frac{\left(1-v^2 \right)^{2n}}{1+v^{4n+2}}\, dv &= \int_{0}^{1} \left(1-v^2 \right)^{2n}\, dv - \int_{0}^{1} \frac{\left(1-v^2 \right)^{2n}v^{4n+2}}{1+v^{4n+2}}\, dv \\ &= \frac{\sqrt{\pi}}{2}\frac{\Gamma(2n+1)}{\Gamma(2n+\frac{3}{2})} +\mathcal{O}(16^{-n}). \end{align*}

Here, the first integral is computed using the beta function identity $\text{(1.2)}$ and the second integral is estimated using the fact that $0 \leq v^2(1-v^2) \leq \frac{1}{4}$ for all $v \in [0,1]$. So it follows that

$$ S_n = \frac{2}{\sqrt{\pi}}\frac{\Gamma(2n+2)}{\Gamma(2n+\frac{3}{2})} + \mathcal{O}\left(\frac{n}{16^n}\right). $$

By the Stirling's approximation, we easily check that $\Gamma(2n+2)/\Gamma(2n+\frac{3}{2}) \sim \sqrt{2n}$ as $n\to\infty$. Here, the relation $\sim$ means that the ratio of two quantities goes to $1$. From this, we obtain

$$ g(n) = \bbox[border:2px dashed green,6px]{ \frac{4}{\pi} \left( \frac{\Gamma(2n+2)}{\Gamma(2n+\frac{3}{2})} \right)^2 + \mathcal{O}\left(\frac{n^{3/2}}{16^n}\right) }. \tag{2.2}$$

This already tells that $g(n) \sim \frac{8}{\pi}n$ as $n\to\infty$, so we can expect that $g(n+1) - g(n) = \frac{8}{\pi}$ if it converges. This is indeed true, as we have

$$ g(n+1) - g(n) = \frac{4}{\pi} \underbrace{ \left( \left( \frac{4n+6}{4n+5}\cdot\frac{4n+4}{4n+3} \right)^2 - 1 \right) }_{\sim \frac{1}{n}} \cdot \underbrace{ \left( \frac{\Gamma(2n+2)}{\Gamma(2n+\frac{3}{2})} \right)^2 }_{\sim 2n} + o(1) $$

The upshot of this calculation is what we have expected:

$$ \lim_{n\to\infty} \left( g(n+1)-g(n) \right) = \frac{8}{\pi}.$$

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