I hope I haven't overseen a thread to this question.
I really struggle with a part of an exercise we have to solve in my Analysis Lecture.
We have a Fourier Series given as:
$$\sum\nolimits_{k=1}^\infty \frac{2(-1)^{k+1}}{k} \sin(kx)$$
And we do have the beginning of a sum, that we are asked to calculate, given as:
$$\frac{1}{1}-\frac{1}{2}+\frac{1}{4}-\frac{1}{5}+\frac{1}{7}-\frac{1}{8}+...$$
So we looked for a value of $x$ and a prefactor that are fullfilling the sum. We ended up with this:
$$\frac{1}{\sqrt{3}}\sum\nolimits_{k=1}^\infty \frac{2(-1)^{k+1}}{k} \sin \left(\frac{k \pi}{3} \right)$$
Now I come to the Main Question:
How can I compute an infinite sum? We have done a computation for $\sum\nolimits_{k=1}^\infty \frac{1}{k²} = \frac{\pi}{6}$ but I couldn't really understand the way our lecturer computed it. I guess the way to solve the sum is very similar to the series we computed in the lecture, though.
So I hope I can get some help here on this stackexchange. I only saw threads that asked for validation of their results, but sadly no further explanation.
Side questions:
- Is the reult of a infinite series its limit?
- Is the fourier series we made up correct to describe the sum?
Thanks in advance for everyone who takes the time to help.