Suppose $f_n$ are continuous real valued functions on $[a, b]$ such that $\sum_{n=1}^\infty f_n$ converges; does it follow that the sequence $a_N =\sum_{n=1}^N f_n$ converges uniformly?
Since $f_n$ is bounded on the compact set $[a,b]$, does uniform convergence follow by the Weierstrass M-test?
No, $\sum_{n=1}^\infty f_n$ may not converge uniformly. For example, take $a= 0$, $b= 1$ and
$f_1 = x$
$f_n = x^n - x^{n-1}$ for $n \geq 2$.
Then $\sum_{n=1}^N f_n = x^N$, and therefore $\sum_{n=1}^\infty f_n$ converges point-wise, but it does not converge uniformly on $[0,1]$.
Also, it is true that each $f_n$ is bounded, being continuous on a compact set. The problem with your argument about using the Weierstrass M-test is that the series of bounds on $f_n$ do not form a convergent series.
You can check that each $|f_n| \leq \frac{(n-1)^{(n-1)}}{n^n} = M_n$ (say), for $n \geq 2$, and you can't find a better bound. But $\sum_{n=1}^\infty M_n$ does not converge.
Hence, the Weierstrass M-test cannot be used here.