Convergence, continuity and differentiability of $f(x)=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}+\ldots$

Question

I am self-learning Real Analysis from the text, Understanding Analysis by Stephen Abbott. I'd like someone to verify, if my below proof and deductions are rigorous and technically correct.

[Abbott 6.4.6] Let

\begin{equation*} f( x) =\frac{1}{x} -\frac{1}{x+1} +\frac{1}{x+2} -\frac{1}{x+3} +\frac{1}{x+4} -\dotsc \end{equation*}

Show that $\displaystyle f$ is defined for all $\displaystyle x >0$. Is $\displaystyle f$ continuous on $\displaystyle ( 0,\infty )$? How about differentiable?

Proof.

Well-definedness of $\displaystyle f$.

Define

\begin{equation*} f_{n}( x) =\frac{( -1)^{n}}{x+n} \end{equation*} Then,

\begin{equation*} f( x) =\sum _{n=0}^{\infty } f_{n}( x) \end{equation*} Let $\displaystyle x_{0}$ be an arbitrary point, such that $\displaystyle x_{0} >0$. Fix $\displaystyle x=x_{0}$.

We have:

\begin{equation*} \frac{1}{x_{0}} \geq \frac{1}{x_{0} +1} \geq \frac{1}{x_{0} +2} \geq \dotsc \geq 0 \end{equation*} Moreover,

\begin{equation*} \lim \frac{1}{x_{0} +n} =0 \end{equation*} By the Alternating Series Test for convergence, $\displaystyle \sum _{n=0}^{\infty } f_{n}( x)$ converges pointwise on $\displaystyle x >0$.

Continuity of $\displaystyle f$.

Let $\displaystyle [ a,\infty )$ be any interval such that $\displaystyle a >0$. Let us group each of pair of terms of $\displaystyle f$ and write:

\begin{equation*} \begin{array}{ c l } f( x) & =\left(\frac{1}{x} -\frac{1}{x+1}\right) +\left(\frac{1}{x+2} -\frac{1}{x+3}\right) +\left(\frac{1}{x+4} -\frac{1}{x+5}\right) +\dotsc \end{array} \end{equation*}

Define:

\begin{equation*} g_{n}( x) =\frac{1}{( x+2n)} -\frac{1}{( x+2n+1)} =\frac{1}{( x+2n)( x+2n+1)} \end{equation*}

Then,

\begin{equation*} f( x) =\sum _{n=0}^{\infty } g_{n}( x) \end{equation*} Now, $\displaystyle g_{0}( x) =\frac{1}{a( a+1)} =M_{0}$. Moreover,

\begin{equation*} 0\leq g_{n}( x) \leq \frac{1}{( 2n)( 2n+1)} \leq \frac{1}{4n^{2}} =M_{n} \end{equation*}

for all $\displaystyle n\geq 1$. Since $\displaystyle \sum _{n=0}^{\infty } M_{n}$ converges, by the Weierstrass $\displaystyle M$-Test, $\displaystyle \sum_{n=0}^{\infty } g_{n}( x)$ converges uniformly on $\displaystyle [ a,\infty )$ for any $\displaystyle a >0$.

Since each $\displaystyle g_{n}( x)$ is continuous for $\displaystyle x >0$, by the Term-by-term continuity theorem, $\displaystyle f( x)$ is continuous on $\displaystyle [ a,\infty )$, where $\displaystyle a >0$. Thus, $\displaystyle f$ is continuous on $\displaystyle ( 0,\infty )$.

Differentiability of $\displaystyle f$.

Since each $\displaystyle g_{n}( x)$ is differentiable for $\displaystyle x >0$, by the Term-by-Term differentiability theorem, $\displaystyle f$ is differentiable on $\displaystyle [ a,\infty )$ where $\displaystyle a >0$. Thus, $\displaystyle f$ is differentiable on $\displaystyle ( 0,\infty )$.

Answer 1

It can be invalid to sum a series by arbitrarily rearranging terms when the series is not absolutely convergent. However, you can sum any convergent series in pairs as you have done even if the series is only conditionally convergent because $f_{2n}(x) \to 0$ as $n \to \infty$ and

$$\sum_{j= 1}^n\{|f_{2j-2}(x)| -|f_{2j-1}(x)|\}=\sum_{k=0}^{2n-1}f_k(x) = \sum_{k=0}^{2n}f_k(x)- f_{2n}(x),$$

so that partial sums with odd and even numbers of terms converge to the same limit.

Nevertheless, this pairing argument is unnecessary. Since $\frac{1}{n+x} \to 0$ as $n \to \infty$ both monotonically and uniformly and $\sum_{k=0}^n (-1)^k$ is uniformly bounded for all $n$ and (trivially) for all $x$, the Dirichlet test for series of functions implies that the series convergence is uniform.

By showing that $\sum f_n'(x)$ is uniformly convergent you can finish the part about differentiability.