Exercise 6.5.2

Question

Find suitable coefficients $\left(a_{n}\right)$ so that the resulting power series $\sum a_{n} x^{n}$ has the given properties, or explain why such a request is impossible.

\begin{enumerate}[label=(\alph*)] 
 \item Converges for every value of $x \in \mathbf{R}$.
 \item Diverges for every value of $x \in \mathbf{R}$.
 \item Converges absolutely for all $x \in[-1,1]$ and diverges off of this set.
 \item Converges conditionally at $x=-1$ and converges absolutely at $x=1$.
 \item Converges conditionally at both $x=-1$ and $x=1$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] \item $a_n = 0$ \item Impossible as $x = 0$ will always converge \item $a_n = \frac{1}{n^2}$. For $x=1$ this converges, while for $x > 1$ the series diverges because $$\frac{x^n}{n^2} <\frac{x^{2n}}{4n^2} \Longleftrightarrow 4 < a^n$$ meaning that once $n > \log_x(4) = \ln(4) / \ln (x)$, the terms will start increasing (whereas they must approach 0 for the series to converge). A similar argument can be made for $x < -1$. \item Impossible because $\left|a_n x^n\right| = \left|a_n (-x)^n\right|$, and substituting $x = 1$ shows that the series at $-1$ is going to be the same as that at $1$ considered absolutely. \item $a_n = 0$ for odd $n$ and $a_n = (-1)^{n/2}/n$ for even $n$. This in effect takes only the even-powered terms of the power series, which are always positive. We then get the alternating harmonic series (scaled by 0.5) in $x^2$ which diverges absolutely but converges conditionally. \end{enumerate}

Exercise 6.5.2

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Exercise 6.5.2

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