Convergence of sine series I'm reading the book Heat Transfer by J.P. Holman. On the chapter of unsteady-state conduction, page 140, the author remarks:

The final series solution is therefore:
  $${\theta(x,t) \over \theta_i} = 
{4\over \pi} \sum^{\infty}_{n=1} {1\over n} e^{-\left({n\pi/L}\right)^2\alpha \,t}\sin{n\pi x \over L}$$
  We note, of course, that at $t=0$ the series on the right side of the Equation must converge to unity for all values of x.

In this equation $0 < x < L$, and $\alpha$ is a finite constant. My question is, how can I proof that
$${4 \over \pi}\sum^{\infty}_{\substack{n=1}} {1\over n} \sin{n\pi x \over L} = 1$$
Additional information: The solution presented above solves the PDE:
$${\partial^2 \theta(x,t) \over \partial x^2} = {1\over \alpha}{\partial^2 \theta(x,t) \over \partial t^2} $$ with initial and boundary conditions:
\begin{align}
\theta(x,0) &= \theta_i \qquad &0\leq x \leq L\\ 
\theta(0,t) &=0  \qquad & t > 0 \\
\theta(L,t) &=0  \qquad & t > 0
\end{align}
 A: You have written a wave equation instead of a heat equation. The equation is likely to be
$$
               \frac{\partial\theta}{\partial t}=\alpha\frac{\partial^2\theta}{\partial x^2}, \\
                 \theta(x,0)=\theta_i \\
                 \theta(0,t)=0,\;\; t > 0, \\
                 \theta(L,t)=0,\;\; t > 0.
$$
The separated solutions $\theta(x,t)=X(x)T(t)$ must satisfy
\begin{align}
                X''(x)=\lambda X(x), \;\; & \;\; T'(t)=\lambda T(t). \\
                X(0)=0,\; X(L)=0. \;\; &
\end{align}
with the added initial condition $\theta(x,0)=\theta_i$ is a constant function. The solutions of the $X$ equations determine $\lambda_n=n^2\pi^2/L^2$, with corresponding solutions $X_n$ that are constant multiplies of $X_n(x)=\sin(n\pi x/L)$. The corresponding solutions $T_n$ are $T_n(t)=e^{-\alpha n^2\pi^2 t/L^2}$. The general solution is then
$$
             \theta(x,t) = \sum_{n=1}^{\infty}A_n e^{-\alpha n^2\pi^2t/L^2}\sin(n\pi x/L),
$$
where the constants $A_n$ are determined by the initial condition
$$
           \theta_i = \sum_{n=1}^{\infty}A_n\sin(n\pi x/L)
$$
and the orthgonality conditions $\int_{0}^{L}X_n(x)X_m(x)dx = 0$ for $n\ne m$:
$$
        \theta_i\int_{0}^{L}\sin(n\pi x/L) = A_n\int_{0}^{L}\sin^2(n\pi x/L)dx \\
    \theta_i \frac{L}{n\pi}\{1-\cos(n\pi)\} = A_n \frac{L}{2} \\
         \frac{2}{n\pi}(1-(-1)^n)\theta_i=A_n.
$$
As you noted, there is a problem at $x=0,L$ for $t=0$ and this is due to the conditions imposed. If you want continuity in $t$ at $t=0$ on $[x=0,x-L]$, then the conditions are incompatible because $\theta(x,0)=\theta_i$ at $x=0$ and $x=L$, which is at odds with $\theta(0,t)=0$ for $t > 0$ and $\theta(L,t)=0$ for $t > 0$. So, unless you have a discontinuity in $t$ at $t=0$ at $x=0,L$, then the solution will not match all the required conditions. This is reflected in the series $\sum_{n=1}^{\infty}A_n\sin(n\pi x/L)$, which converges to $0$ at $x=0$ and at $x=L$, but converges to $\theta_i$ in $0 < x < L$. The peculiar behavior should mostly be blamed on the conditions for the problem, which require a discontinuity in $t$ at $t=0$ for the enpoints $x=0$ and $x=L$.
The solution given in the problem is incorrect. And you cannot prove the identity you want because the sin series of the constant function is unique, and I have given the correct one. And, that series I have given will converge to $\theta_i$ on $(0,L)$, but of course converges to $0$ at $0,L$.
A: Regards..Sergio. One of ways to see the proof is by understanding the way the exact general solution is achieved. This will also make you understand more about the PDE. 
One other way is to integrate the series. By writing the convergence : 
$$     \sum_{n=1}^{\infty}  \frac{1}{n \pi } \sin \left(\frac{n \pi x}{L} \right)  = c $$ where $c$ is the convergence value.
After this you should first integrate the sine series :
$$ \int \sum_{n=1}^{\infty} \frac{1}{n \pi } \sin \left(\frac{n \pi x}{L} \right) dx =
\sum_{n=1}^{\infty} - \left(  \frac{L}{(n \pi)^2 } \right) \cos \left( \frac{n \pi x}{L} \right) = c L $$
Along the interval $ 0<x<L $. You will get the form below :
$$  \left(  \frac{2L}{( \pi)^2 } \right)  \sum_{k=0}^{\infty} \frac{1}{(2k+1)^2} = cL $$
From there the series below will be useful :
$$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi}{6} $$
To find the value of :
$$ \sum_{k=0}^{\infty} \frac{1}{(2k + 1)^2}  $$
to find the value $c$. But i found that the value is different from what you post.
Hope this would be useful. 
