求x使∑(Hₙsin nx)/n收敛

hbghlyj

Mathematical Analysis, Apostol著, 213页, Exercise 8.26

Determine all real values of $x$ for which the following series converges: $$\sum_{i=1}^∞\left(1+\frac12+\dots+\frac1n\right)\frac{\sin nx}n.$$

If $\sin\frac x2 = 0,$ then the sum is obviously zero. So assume otherwise. Then, show two things:

• $b_n = \sin x+\sin(2x)+...+\sin(nx)$ is uniformly bounded over all $n$ for given $x$ by Theorem 8.30.
• $a_n = \frac 1n\left(1+\frac 12 +\frac 13+\dots + \frac 1n\right)$ is monotonically decreasing to zero.

  (MSE)Proof: Use the fact that $x↦\frac1x$ is decreasing over $[1,∞)$ to get $$ 0<1+\frac12+\frac13+\cdots+\frac1n<1+\int_1^n\frac{dx}x=1+\log n $$giving $$ 0<\frac1n\left(1+\frac12+\frac13+\cdots+\frac1n\right)<\frac1n+\frac{\log n}n. $$

Then, you will be ready to use Theorem 8.28 (Dirichlet Criterion).

hbghlyj

To show that the series does not converge absolutely, it suffices to show that $\sum\nolimits_{n = 1}^\infty  {|\frac{{\sin n}}{n}|}  = \infty$
math.stackexchange.com/questions/1204593/confused-about-switchin ... tegrals?noredirect=1
$$\int_0^{\infty}\frac{|\sin x|}{x}=\sum_{n=1}^{\infty}\int_{(n-1)\pi}^{n \pi}\frac{|\sin x|}{x}\ dx\geq \sum_{n=1}^{\infty}\frac{1}{n\pi}\int_0^{\pi}|\sin x|\ dx=\infty$$

hbghlyj

这个帖子是$x=1$的情况:
$$\sum_{n=1}^{\infty} \frac{\sin(n)}{n} \cdot \left(1+\frac{1}{2} + \cdots + \frac{1}{n}\right)$$

math.stackexchange.com/questions/370402/series-sum-frac-sinnn-cd ... frac1n-right-converg

The series is convergent and, for fun, I'll provide a derivation.

Let's first rewrite your sum as ($H_n$ is the $n$-th harmonic number) :
$$\tag{1}S:=\sum_{n=1}^{\infty} H_n\frac{\sin(n)}{n}=\Im\left(\sum_{n=1}^{\infty} H_n\frac{e^{in}}{n}\right)$$

Since $H_n\sim \log(n)\,$ as $\,n\to\infty$ (because the 'harmonic integral' corresponding to the arithmetic sum is the logarithm)

To evaluate $S$ let's use generating functions and suppose that the derivative of $f$ is defined by :
$$\tag{2}f'(x):=\sum_{n=1}^{\infty} H_n\;x^n=-\frac 1{1-x}\ln(1-x)$$
(expand $\frac 1{1-x}=1+x+x^2+\cdots\ $ and $\ -\ln(1-x)=x+\frac {x^2}2+\frac{x^3}3+\cdots$ and multiply !)

Integrating $(2)$ we get :
$$\tag{3}f(x)=\sum_{n=1}^{\infty} H_n \frac{x^{n+1}}{n+1}=-\int\frac{\ln(1-x)}{1-x} dx=\frac 12(\ln(1-x))^2$$
so that (using $\;H_n=H_{n+1}-\frac 1{n+1}\;$ and setting $\,k:=n+1\;$) :
$$\sum_{k=2}^{\infty} H_k \frac{x^k}k-\sum_{k=2}^{\infty} \frac{x^k}{k^2}=\sum_{k=1}^{\infty} H_k \frac{x^k}k-\sum_{k=1}^{\infty} \frac{x^k}{k^2}=\frac 12(\ln(1-x))^2$$
$\operatorname{Li}_2(x):=\sum_{k=1}^{\infty} \frac{x^k}{k^2}$ is the dilogarithm  but we prefer to use directly :
$$\tag{4}\sum_{k=1}^{\infty} H_k \frac{x^k}k=\sum_{k=1}^{\infty} \frac{x^k}{k^2}+\frac 12(\ln(1-x))^2$$
For $x=e^i$ in $(1)$ we get following expression of $S$ :
\begin{align}
S&=\Im\left(\sum_{k=1}^{\infty} H_k \frac{e^{ik}}k\right)\\
S&=\sum_{k=1}^{\infty} \frac{\sin(k)}{k^2}+\frac 12\Im\left(\ln\bigl(1-e^i\bigr)\right)^2\\
S&=\boxed{\displaystyle\operatorname{Cl}_2(1)-\ln\left(2\sin\left(\frac 12\right)\right)\frac{\pi-1}2}\\
S&\approx 1.05895346485231034922735
\end{align}
with $\operatorname{Cl}_2$ the Clausen function  : $\displaystyle\operatorname{Cl}_2(x)=-\int_0^x \ln\left(2\sin\left(\frac t2\right)\right)dt$