I trying to prove this statement but have not succeed. I tried to use the Weierstrass $M$-Test, but couldn't get very far. The question is as follows:
For all $r>0$, prove that the series
$\sum_{n=1}^\infty \left(\sin^8(n^3x)\over n^5+x^2\right)'$ is uniformly convergent on the interval $[-r,r]$
Basically, I tried to find a sequence that is bigger than my function for the $M$-Test. First, I took the derivative (to simplify, let the function be $f_n(x)$):
$f_n'(x)={8\sin^7(n^3x)\cos(n^3x)n^3[n^5+x^2]-\sin^8(n^3x)2x\over (n^5+x^2)^2}$
$\left|f_n'(x)\right|=\left|{8\sin^7(n^3x)\cos(n^3x)n^3[n^5+x^2]-\sin^8(n^3x)2x\over (n^5+x^2)^2}\right|$
Using the triangle inequality:
$\left|f_n'(x)\right|=\left|{8\sin^7(n^3x)\cos(n^3x)n^3[n^5+x^2]-\sin^8(n^3x)2x\over (n^5+x^2)^2}\right|$
$\left|f_n'(x)\right|\leq\left|{8\sin^7(n^3x)\cos(n^3x)n^3[n^5+x^2]\over (n^5+x^2)^2}\right|+\left|{-\sin^8(n^3x)2x\over (n^5+x^2)^2}\right|$
Since both $\sin(x)$ and $\cos(x)$ are always less or equal to 1 in absolute value, by substuting they for 1, we achieve a function that is bigger or equal to the right side of the inequality, and therefore, bigger or equal to $\left|f_n'(x)\right|$:
$\left|f_n'(x)\right|\leq\left|{8n^3[n^5+x^2]\over(n^5+x^2)^2}\right|+\left|{2x\over (n^5+x^2)^2}\right|$
$\left|f_n'(x)\right|\leq\left|{8n^3\over n^5+x^2}\right|+\left|{2x\over (n^5+x^2)^2}\right|$
We can check that the first term achieves his maximum at $x=0$. So lets focus on the second one for now:
$g_n(x)={2x\over (n^5+x^2)^2}$
$g_n'(x)={2(n^5+x^2)^2-2x\cdot2(n^5+x^2)\cdot2x\over (n^5+x^2)^4}$
$g_n'(x)={2(n^5+x^2)-2x\cdot2\cdot2x\over (n^5+x^2)^3}$
$g_n'(x)={2n^5+2x^2-8x^2\over (n^5+x^2)^3}$
$g_n'(x)={2n^5-6x^2\over (n^5+x^2)^3}$
Setting the derivative equal to zero gives the critical points of $g_n(x)$:
$0={2n^5-6x^2\over (n^5+x^2)^3}$
$0=2n^5-6x^2$
$x=\sqrt{n^5\over3}$
Since the denominator is always positive, the sign of $g_n'$ will be the same of the numerator. If $x<\sqrt{n^5\over3}$, then $g_n'>0$. If $x>\sqrt{n^5\over3}$, then $g_n'<0$. Therefore, $\sqrt{n^5\over 3}$ is a maximum of $g_n$. Then:
$|g_n(x)|\leq \left|g_n\left(\sqrt{n^5\over3}\right)\right|$
$|g_n(x)|\leq \left|9\sqrt{n^5\over3}\over 8n^{10}\right|$
$|g_n(x)|\leq \left|9\sqrt{n^5\over3}\over 8n^{10}\right| < \left|9\sqrt{n^6}\over 8n^{10}\right|=\left|9n^3\over 8n^{10}\right|={9\over 8n^{7}}$
Therefore:
$|f_n'(x)|\leq\left|{8n^3\over n^5+x^2}\right|+\left|{2x\over (n^5+x^2)^2}\right|$
$|f_n'(x)|\leq {8n^3\over n^5}+{9\over 8n^{7}}$
$|f_n'(x)|\leq {8\over n^2}+{9\over 8n^{7}}$
$|f_n'(x)|\leq {8\over n^2}+{9\over 8n^{7}}\leq {8\over n^2}+{2\over n^{2}}$
$|f_n'(x)|\leq {10\over n^2}$
Since $\sum_{n=1}^\infty {10\over n^2}$ converges, then $\sum_{n=1}^\infty f_n'(x)$ converges uniformly.
So, apparently I proved that the derivative converges uniformly for all values of x. But, since I did not use the [-r,r] interval on the proof, I guess that my proof is wrong.