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There was a book problem I recently solved going like this:

$\sum_{n=0}^\infty a_n^2x^n$ converges on $[-6,6]$

What is the largest interval $\sum_{n=0}^\infty a_nx^n$ is guaranteed to converge on?


I already understand this:

$\limsup_{n\rightarrow\infty}|a_n^2|^{1/n}=1/6$

$\Rightarrow\limsup_{n\rightarrow\infty}|a_n|^{1/n}=1/\sqrt6$

$\Rightarrow$Radius of Convergence is $(-\sqrt6,\sqrt6)$ for $\sum_{n=0}^\infty a_nx^n$


What I was wondering however, is what are examples of what the values for $a_n$ could be where the endpoints dont converge for the series, but the series with $a_n^2$ converges on [-6,6]?

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  • $\begingroup$ Can you think of something if I tell you to consider $a_n = \frac{1}{n}\cdot b_n$? $\endgroup$ Commented May 1, 2017 at 20:19
  • $\begingroup$ The first = should be $\le.$ $\endgroup$
    – zhw.
    Commented May 1, 2017 at 20:56

1 Answer 1

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Let $R$ be the radius of convergence of the power series

$$\sum_{n = 0}^{\infty} a_n x^n.\tag{1}$$

Then define $c_n = a_n\cdot R^n$. We want the series

$$\sum_{n = 0}^{\infty} a_n^2 x^n\tag{2}$$

to converge on the closed interval $[-R^2,R^2]$, and that is the case if and only $(2)$ converges for $x = R^2$ and $x = -R^2$. If we consider only real coefficients $a_n$, then $a_n^2 \geqslant 0$ for all $n$, and the convergence of $(2)$ for $x = R^2$ already implies the (absolute and uniform) convergence on the interval $[-R^2,R^2]$. Plugging $x = R^2$ into $(2)$, we want the convergence of

$$\sum_{n = 0}^{\infty} a_n^2 R^{2n} = \sum_{n = 0}^{\infty} c_n^2.$$

And we want the series $(1)$ to diverge for $x = R$ and for $x = -R$. Plugging those values in, we want the divergence of

$$\sum_{n = 0}^{\infty} a_n R^n = \sum_{n = 0}^{\infty} c_n\qquad\text{and}\qquad \sum_{n = 0}^{\infty} a_n (-R)^n = \sum_{n = 0}^{\infty} (-1)^n c_n.$$

So we look for a sequence $(c_n)$ such that $\sum c_n$ and $\sum (-1)^n c_n$ diverge, but $\sum c_n^2$ converges. If we don't care about $\sum (-1)^n c_n$, then a famous example of such a sequence is $\bigl(\frac{1}{n+1}\bigr)_{n\in\mathbb{N}}$. Unfortunately, the alternating harmonic series converges, so we cannot use just that sequence. But we can modify that sequence to reach our goal.

One easy way is to nullify the effect of the $(-1)^n$ by inserting zeros into the sequence, say $c_n = 0$ for odd $n$ and $c_n = \frac{1}{n+1}$ for even $n$.

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