Exercise 6.4.5

Question

\begin{enumerate}[label=(\alph*)] 
  \item Prove that
    $$
    h(x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n^{2}}=x+\frac{x^{2}}{4}+\frac{x^{3}}{9}+\frac{x^{4}}{16}+\cdots
    $$
    is continuous on $[-1,1]$.
  \item The series
    $$
    f(x)=\sum_{n=1}^{\infty} \frac{x^{n}}{n}=x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+\cdots
    $$
    converges for every $x$ in the half-open interval $[-1,1)$ but does not converge when $x=1$. For a fixed $x_{0} \in(-1,1)$, explain how we can still use the Weierstrass M-Test to prove that $f$ is continuous at $x_{0}$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
    \item For $x \in [-1, 1]$, we have
    $$\left|\frac{x^n}{n^2}\right| \leq \frac{1}{n^2} = M_n $$
    and since $\sum \frac{1}{n^2}$ converges (Example 2.4.4), $h$ converges uniformly and is therefore continuous.
    \item Given a fixed $x_0$, we can consider the interval $(-a, a) \subset [-1, 1)$ where $-1 < -a < \left|x_0\right| < a < 1$. Then by setting $M_n = \frac{a^n}{n}$ we will have $M_n > \frac{x_0^n}{n}$ in a neighbourhood around $x_0$, allowing us to show via the M-Test that $f$ is continuous at $x_0$.
 \end{enumerate}

Exercise 6.4.5

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Exercise 6.4.5

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