Wolfram Alpha says
Sum[Sin(Pi*n/4)]/(Pi*n/4),{n,-Infinity,Infinity}]
is equal to 4 but I don't know how to resolve it...
In my signal and system homework,this sequence is required to be identify what the type of this signal is,such as energy signal,power signal. So I encounter this problem.
Even though the n=0 is not defined,the W|A also says
Sum[Sin(Pi*n/4)]/(Pi*n/4),{n,1,Infinity}]
is 1.4999999999...(The symbolic exptession is very complex,it can not simplify to 1.5).
Or,Can we prove the summation have boundry?
Howerver,Using the Fourier transform,the Integrate[Sin(x)/x,{n,-Infinity,Infinity}] can be resolve.But if it is a sequence,How to solve it?
(need to fix the end. This is the right idea, but the end conclusion appears wrong.)
$$-\log(1-z) = \sum_{n=1}^\infty \frac{z^n}{n}$$
when $|z|\leq 1$ and $z\neq 1$.
Then $\sum_{n=1}^\infty \frac{\sin nx}{nx}$ is the imaginary part of:
$$\frac{1}{x}\sum_{n=1}^\infty \frac{e^{inx}}{n}=-\frac{\log(1-e^{ix})}{x}$$
When $x=\frac{\pi}{4}$, this is the imaginary part of:
$$-\frac{\log\left(1-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2}\right)}{\frac{\pi}{4}}$$ The imaginary part of this is $$\frac{\arctan(\sqrt 2+1)}{\frac{\pi}{4}}$$
and $\arctan(\sqrt 2+1)=\frac{3\pi}{8}$.
To see that last, take the triangle $A=0,B=1,C=1-e^{ix}$. Since $|AB|=|BC|=1$, you have an isosceles triangle. The imaginary part of $-\log\left(1-e^{ix}\right)$ is just the angle $\angle A$. $\angle B=x=\frac{\pi}{4}$, and so $\angle A=\angle C$ gives $\angle C=\frac{3\pi}{8}$.