I need help in computing this infinite series $\displaystyle \sum_{k=1}^{\infty}\dfrac{(-1)^{k}4(2k-1)}{\pi[4(2k-1)^2-1]}$. What value of $t$ do I use? The Fourier series I've been given is: $$ \frac{1}{2}\cos(t)-\sum_{k=1}^{\infty} \dfrac{4k}{\pi \left(4k^2-1\right)} \sin(2kt) $$

  • $\begingroup$ Not sure what $t$ has to do with it? $\endgroup$ – gt6989b Oct 13 '15 at 16:00
  • $\begingroup$ Hi, in my question it says Using the Fourier series by taking a proper value for t in this Fourier series, compute the sum of the following infinite series. $\endgroup$ – Andrew Cavagnino Oct 13 '15 at 16:01
  • $\begingroup$ You should be able to see immediately that where you have "sin(2kt)" in the Fourier series, you have (-1)^k in the second. That is, of course, 1 if k is even and -1 if k is odd. For what values is sin equal to -1 and 1? What values of t give those for "2kt"? $\endgroup$ – user247327 Oct 13 '15 at 16:17

$$\frac{1}{2}\cos(t)-\sum_{k\geq 1}\frac{4k}{\pi(4k^2-1)}\sin(2k t) \tag{1}$$ is the Fourier series over $(-\pi,\pi)$ of a function that equals $\cos(t)$ over $(-\pi,0)$ and $0$ over $(0,\pi)$.

By evaluating $(1)$ at $t=\frac{\pi}{4}$, we have: $$ -\sum_{k\geq 1}\frac{4k}{\pi(4k^2-1)}\sin\left(\frac{\pi k}{2}\right) = -\frac{1}{2\sqrt{2}}\tag{2} $$ but $\sin\left(\frac{\pi k}{2}\right)$ equals $0$ if $k$ is even and $\pm 1$ if $k$ is odd, hence we may write $(2)$ as: $$ \sum_{k\geq 1}\frac{4(2k-1)}{\pi(4(2k-1)^2-1)}(-1)^k = -\frac{1}{2\sqrt{2}}.\tag{3}$$

Another approach for computing the last series is the following:

$$\begin{eqnarray*} \sum_{k\geq 0}\left(\frac{1}{4k+1}+\frac{1}{4k+3}\right)(-1)^k &=& \sum_{k\geq 0}(-1)^k \int_{0}^{1}\left(x^{4k}+x^{4k+2}\right)\,dx\\&=&\int_{0}^{1}\sum_{k\geq 0}(-1)^k\left(x^{4k}+x^{4k+2}\right)\,dx\\&=&\int_{0}^{1}\frac{1+x^2}{1+x^4}\,dx.\tag{4}\end{eqnarray*}$$

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    $\begingroup$ How does one guess the function associated to the given fourier series? $\endgroup$ – Calvin Khor Oct 13 '15 at 16:38
  • $\begingroup$ @CalvinKhor: by experience. $\endgroup$ – Jack D'Aurizio Oct 13 '15 at 16:41
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    $\begingroup$ Was afraid so. Thanks anyway :) $\endgroup$ – Calvin Khor Oct 13 '15 at 16:45

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