On the use of Weierstrass' M-test for uniform convergence of series including unbounded terms

Question

Let $A$ be a subset of $\mathbb{R}$ and for each integer $k\in\mathbb{N}$ consider a sequence of functions $\{f_k(x)\}_{k=1}^\infty$ defined on the set $A$. Suppose that there is an integer $n^*$ such that $\sup_{x\in A}|f_k(x)|\leq M_k$ for every $k>n^*$ and that $\sum_{k=n^*+1}^\infty M_k<\infty$. Hence by the Weierstrass M-test the series $\sum_{k=n^*+1}^\infty f_k(x)$ converges uniformly (and absolutely) on the set $A$.

Now if all of (or, some of) the functions $\{f_k(x)\}_{k=1}^{n^*}$ are unbounded on the set $A$, can we still say that the whole series $\sum_{k=1}^\infty f_k(x)$ converges uniformly (and absolutely) on the set $A$ ? I saw in many comments in this site (when answering questions related to the use of the Weierstrass M-test) saying that when we remove unbounded first finite terms from the series and apply the Weierstrass M-test, if the rest of the series is uniform convergent (by the Weierstrass M-test), then the whole series is uniform convergent, too. If so, how can we arrive at this result? (Because, we might down to the case $\infty-\infty$ when removing unbounded terms.)

(By the way, in this case, the Cauchy Criterion would be sufficient to conclude that the whole series is uniformly convergent on the set $A$. Is this correct?)

Answer 1

Suppose $ \sum\limits_{k=n^{*}+1}^{\infty} f_k(x)$ converges uniformly to $G(x)$. Let $F(x)= \sum\limits_{k=1}^{n^{*}}f_k(x)$. Consider $| \sum\limits_{k=1}^{N} f_k(x)-(F(x)+G(x))|$ where $N >n^{*}$. This is same as $| \sum\limits_{k=n^{*}+1}^{N} f_k(x)-G(x)|$ because the first $n^{*}$ terms simply cancel out. Now just apply definition of uniform convergence of $ \sum\limits_{k=n^{*}+1}^{\infty} f_k(x)$ to $ G(x)$ to complete the proof.