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Assume $n^4 \lvert c_n \rvert \le M$, for all $n=1,2,....$ Define the infinite series $$y(x,t) = \sum_{n=1}^\infty c_n \sin(nx)\cos(nt),$$ (a) Prove that the series is twice differentiable with respect to x and t.

(b) Prove that the series satisfies the wave equation $y_{tt}=y_{xx}$.

What I got so far.

So to prove that the infinite series is twice differentiable with respect to x and t we have to show that the second partial derivatives with respect to x and t are uniformly convergent, i.e. that $$ y_{tt}(x,y) = \sum_{n=1}^\infty c_n \partial_{tt}\sin(nx)\cos(nt) $$ and $$y_{xx}(x,y) = \sum_{n=1}^\infty c_n \partial_{xx}\sin(nx)\cos(nt) $$ are both uniformly convergent.

Turns out that $$y_{tt}(x,y)=y_{xx}(x,y)= \sum_{n=1}^\infty -n^2c_n \sin(nx)\cos(nt)$$ so showing that either $y_{tt}$ or $y_{xx}$ is uniformly convergent implies that the other is as well. Now by the Weierstrass M-test,

(i)$\lvert -n^2c_n \sin(nx)\cos(nt) \rvert \le \lvert n^2 c_n \rvert = n^2 \lvert c_n\rvert =\frac{n^4 \lvert c_n \rvert}{n^2} \le \frac{M}{n^2} $

(ii) The original series is uniformly convergent if $$ \sum_{n=1}^\infty \frac{M}{n^2}$$ is convergent. This is where I run into problems though. Is $M$ a constant? Can I pull it out of the sum and conclude that the series is convergent by the p-series test?

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    Yes you can. It is given to you that $|c_n| \leq \frac{M}{n^4}$ for some constant $M$ and this $M$ is a fixed number that does not depend on $n$. – 2017-02-13

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The M-test says that if for some sequence $M_n,$ we have $|f_n(x)| < M_n$ ($\forall x$) and $\sum_n M_n < \infty$ then $f_n$ is uniformly convergent. You've shown that $|f_n|