Fourier Series Problem and Convergence We are given the following function $f(t)$, which is odd and  periodic with $2π$ in the interval $[0, π]$  is given as,
$$f(t)=\frac{8}{π}\cdot(πt-t^2)$$
i) Show that the f has the fourier serie,
$$\sum _{ n=1 }^{ \infty  }{ \frac { \sin((2n-1)t) }{ (2n-1)^{ 3 } }  } $$
Hint: For any real value of $n$, we know that,
$$\int t\sin(nt)dt=\frac{\sin(nt)}{n^2}-t\frac{\cos(nt)}{n}$$
$$\int  t^{ 2 }\sin(nt)dt=2\frac { \cos(nt) }{ n^{ 3 } } +2t\frac { \sin(nt) }{ n^{ 2 } } -t^{ 2 }\frac { \cos(nt) }{ n}$$
iii) Is the fourieseries convergent with the sum of $f(t)$ for all $t$?
iv) Show, that
$$\sum _{ n=1 }^{ \infty  }{ \frac { (-1)^{ n-1 } }{ (2n-1)^3 }  }=\frac{\pi^3}{32}$$
My Approach
For i, I know that the function is odd therefore $a_n$=0, and hence we only have to find $b_n$. I used the following formula to find the formula of $b_n$.
$$b_n=\frac{2}{π}\cdot\int _{ 0 }^{π  }{f(t)\cdot \sin(nt)dt } $$
But I got the following expression,
$$\frac { 1 }{ 4 } \cdot \frac { nπ\sin(nπ)+2(\cos(nπ)-1) }{ n^{ 3 } } $$
Then to find the riemman expression, I used the following formula,
$$\sum _{ n=1 }^{ \infty  }{ (\frac { 1 }{ 4 } \cdot \frac { nπ\sin(nπ)+2(\cos(nπ)-1) }{ n^{ 3 } } )\cdot \sin(nt) } $$
I have been stuck at this problem for very long now, and I can not seem to understand how I can change that expression into the the one given.
For ii I tried using the ratio test but that didnt given any good result. How other can i determine the convergence? is there a formula for fourier series?
 A: (i) Since $f$ is odd, $b_n = {2 \over \pi} \int_0^{\pi} f(t) \sin (nt) dt = {32 \over \pi^2 n^3} (1-(-1)^n)$.
To see this, let 
$F(t)={8 \over\pi  {n}^{3}}
(- 2 n t \sin( n t)
+\pi  n \sin( n t)
+{n}^{2} {t}^{2} \cos( n t)
-\pi  {n}^{2} t \cos( n t)
-2 \cos( n t) )$,
and note that $F' = f$ for $t \in [0, \pi]$.
We see that $b_n = \begin{cases} {64 \over \pi^2 n^3}, & n \text{ odd} \\
0, & \text{otherwise}\end{cases} $.
Hence the Fourier series is $\phi(t) = \sum_{n=1}^\infty b_{2n-1} \sin ((2n-1)t) = {64 \over \pi^2 } \sum_{n=1}^\infty {\sin ((2n-1)t) \over (2n-1)^3 }$
(ii) This is missing :-).
(iii) Since $f$ is Lipschitz continuous, the Fourier series converges
uniformly (and hence pointwise).
As an aside, there is a nice result called Carleson's theorem which shows that
for a square integrable $f$, the Fourier series converges to $f(t)$ for
almost every $t$. Not applicable here, but indicates that convergence is
generic in some sense.
Note: Showing that the series converges is straightforward (since $|b_n| \le {K \over n^3}$, we can use the comparison test), but that is not the question. The question is to show that
it converges to $f(t)$.
(iv) $\phi({ \pi \over 2}) = {64 \over \pi^2 } \sum_{n=1}^\infty {(-1)^{n-1} \over (2n-1)^3 } = f({ \pi \over 2}) = {8 \over \pi} ({\pi^2 \over 2} - {\pi^2 \over 4}) = {2 \pi}$, from which the result follows.
