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**Opalg** In the last line of the proof, you can't apply the M-test in that way, because $\displaystyle a_n$ depends on $\displaystyle x$. The M-test only applies if $\displaystyle (a_n)$ is a sequence of constants. If the function $\displaystyle f_N(x)$ is bounded, say $\displaystyle |f_N(x)|\leqslant K$ for all $\displaystyle x\in X$, then you can take $\displaystyle a_n = c^{n-N}K$. In that case, the M-test does apply and the conclusion is correct. In particular, if the space $\displaystyle X$ is compact, then $\displaystyle f_N(x)$ will automatically be bounded and the result is correct.

But suppose for example that $\displaystyle X=\mathbb{R}$ and $\displaystyle f_n(x) = 2^{-n}x$. Then $\displaystyle \sum_{n=1}^\infty f_n(x) = x$ for all $\displaystyle x\in\mathbb{R}$, but the convergence is not uniform.