Answer: Alternating Series Beyond the M-Test

Key Idea / Intuition

The Weierstrass M-test is a sufficient condition for uniform convergence, not a necessary one. Alternating series enjoy a much more delicate cancellation mechanism — adjacent terms nearly cancel, and the error after $N$ terms is bounded by the size of the next term, which can be made uniformly small even when the individual terms are not summable. This example cleanly separates the concepts of "absolute uniform convergence" from "uniform convergence via cancellation."

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Formal Proof / Solution

Part (a): Uniform Convergence via the Alternating Series Test (Uniform Version)

Write the partial sums $S_N(x) = \sum_{n=1}^N \frac{(-1)^n}{n+x}$.

The standard Alternating Series Estimation says: if $a_n(x)$ is decreasing to $0$ for each $x$ and the signs alternate, then

$|f(x) - S_N(x)| \leq a_{N+1}(x).$

Here $a_n(x) = \frac{1}{n+x}$. For each fixed $x \in [0,1]$ this is indeed decreasing in $n$ and tends to $0$. Moreover,

$\sup_{x \in [0,1]} a_{N+1}(x) = \sup_{x \in [0,1]} \frac{1}{N+1+x} = \frac{1}{N+1} \xrightarrow{N \to \infty} 0.$

Therefore

$\sup_{x \in [0,1]} |f(x) - S_N(x)| \leq \frac{1}{N+1} \to 0,$

which is uniform convergence.

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Part (b): Continuity of $f$

Each partial sum $S_N(x) = \sum_{n=1}^N \frac{(-1)^n}{n+x}$ is a finite sum of continuous functions on $[0,1]$, hence continuous. Since $S_N \to f$ uniformly, and the uniform limit of continuous functions is continuous, $f$ is continuous on $[0,1]$.

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Part (c): Why the Weierstrass M-Test Fails Here (and Why That's Fine)

The Weierstrass M-test requires finding constants $M_n$ with $\left|\frac{(-1)^n}{n+x}\right| \leq M_n$ for all $x$ and $\sum M_n < \infty$.

The natural bound is $M_n = \frac{1}{n+0} = \frac{1}{n}$, but $\sum \frac{1}{n} = \infty$. Any other bound $M_n$ would have to satisfy $M_n \geq \frac{1}{n+1}$, so no summable dominating sequence exists.

But the M-test is only sufficient, not necessary. It tests for absolute uniform convergence:

$\text{M-test succeeds} \iff \sum_n \sup_x |f_n(x)| < \infty.$

Our series converges uniformly by cancellation between consecutive terms, not by absolute dominance. The correct tool is the Dirichlet/Abel test for uniform convergence (or the uniform alternating series test used above).

Moral: $\sum f_n$ can converge uniformly even when $\sum |f_n|$ diverges uniformly. Cancellation is a genuine and powerful phenomenon in analysis, and the M-test is too blunt an instrument to see it.

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Bonus Remark

In fact $f(x)$ has a closed form. Writing the series as a difference of harmonic series one can show

$f(x) = \sum_{n=1}^\infty \frac{(-1)^n}{n+x} = \frac{1}{2}\left[\psi\!\left(\frac{x+2}{2}\right) - \psi\!\left(\frac{x+1}{2}\right)\right],$

where $\psi$ is the digamma function — confirming continuity and giving a rich connection to special functions.