Using Fourier-series expansion of $f(\theta) = \theta$ one can show that:
$$ \sum_{n=1}^{\infty} \dfrac{(-1)^{n+1} }{n} \sin (2n) = 1. $$
Is there any other way to prove this result? In particular, is there an elementary way to knock this down?
Asked
Active
Viewed 136 times
-1
-
1I would call it "elementary" to use a Fourier series. There's not too many simpler ways to evaluate sums. – K.defaoite Feb 13 '25 at 04:46
-
@K.defaoite The word "elementary" is often used to mean "avoiding complex numbers", which a Fourier series does not. – Misha Lavrov Feb 13 '25 at 05:08
-
Hint: $\sin(2n)=\Im e^{i2n}$, then maybe use $\text{Li}_1(z)$ – Quý Nhân Feb 13 '25 at 05:39
-
regarding the hint, im not sure why you would use $\textrm{Li}_1(z)$ instead of just the series expansion for $\ln (1+z)$ – Aniruda Suswaram Feb 13 '25 at 06:05
-
1Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. See also the section on titles in How to ask a good question. (The part entitled "Make your title your question" is especially relevant to this.) – Martin Brandenburg Feb 13 '25 at 06:52
-
1See https://math.stackexchange.com/a/3361622/42969 for a very similar question. – Martin R Feb 13 '25 at 07:24
1 Answers
2
Let $ n\geq k\geq 1 $. We have the following : $$ \int_{\frac{\pi}{2}-1}^{\frac{\pi}{2}}{2\cos{\left(2kx\right)}\,\mathrm{d}x}=\frac{\left(-1\right)^{k}\sin{\left(2k\right)}}{k} $$
Thus : \begin{aligned}\sum_{k=1}^{n}{\frac{\left(-1\right)^{k+1}\sin{\left(2k\right)}}{k}}&=-\int_{\frac{\pi}{2}-1}^{\frac{\pi}{2}}{2\sum_{k=1}^{n}{\cos{\left(2kx\right)}}\,\mathrm{d}x}\\ &=-\int_{\frac{\pi}{2}-1}^{\frac{\pi}{2}}{\left(\frac{\sin{\left(\left(2n+1\right)x\right)}}{\sin{x}}-1\right)\mathrm{d}x}\\ &=1-\int_{\frac{\pi}{2}-1}^{\frac{\pi}{2}}{\csc{x}\sin{\left(\left(2n+1\right)x\right)}\,\mathrm{d}x}\end{aligned}
Now, $ \csc $ is continuous on $ \left[\frac{\pi}{2}-1,\frac{\pi}{2}\right] $, therefore $ \int_{\frac{\pi}{2}-1}^{\frac{\pi}{2}}{\csc{x}\sin{\left(\left(2n+1\right)x\right)}\,\mathrm{d}x}\underset{n\to+\infty}{\longrightarrow}0 $. (by Riemann–Lebesgue lemma)
Hence $\cdots $
CHAMSI
- 9,222