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

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?)

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.

• Thanks for clarification. Jun 4, 2020 at 12:31