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Given the series $$\sum_{n=0}^{\infty}(-1)^nx^{2^n}$$ determine the radius of convergence, and what can we say when $x=R$ and $-R$?

  1. Is it a power series? Power series should have the form of $$\sum_{n=0}^{\infty}a_nx^n$$ but the given series does not match this form. If not a power series, why can we say about its radius of convergence?

  2. By the ratio test, I get that this series converges when $|x|<1$, diverges when $|x|>1$, so $R=1$, is that right?

  3. When $x=1$ or $-1$, series both becomes $$\sum_{n=0}^{\infty}(-1)^n,$$ then obviously, series diverges. Right?

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    $\begingroup$ It is a power series because $a_n = 0$ when n isn't a power $\endgroup$
    – user217285
    Jul 14, 2015 at 6:01
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    $\begingroup$ $\sum \left(-1\right)^n x^{2^n} = x - x^2 + x^4 - x^8 + x^{16} - \cdots$, so this is clearly a power series. Just plug in some $0\cdot x^n$ terms for the powers of $x$ that do not appear explicitly $\endgroup$
    – fonini
    Jul 14, 2015 at 6:05
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    $\begingroup$ About 3., you are right, except that when $x=-1$, the sum is $-1-1+1-1+1-1+\cdots$. The first term is the only one with odd power of $x$, so its sign doesn't follow the rest of the pattern. $\endgroup$
    – fonini
    Jul 14, 2015 at 6:06

1 Answer 1

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(1): see the comments.

(2): the radius of convergence $\rho$ of $\sum a_nx^n$ is defined to be:

$$ \rho:=\frac{1}{\alpha} $$

where

$$ \alpha = \lim_{n\rightarrow \infty} \sup \sqrt[n]{|a_n|} $$.

Here we have:

$|a_n| = 1$ if $n$ is a power of 2, $|a_n|=0$ otherwise. Thus $\rho=1$.

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  • $\begingroup$ Actually, numbers $2^n$ are not normally what we call "square numbers". But the argument is still OK using "power of two" instead. $\endgroup$
    – GEdgar
    Jul 14, 2015 at 22:13
  • $\begingroup$ @GEdgar Thanks, edited! $\endgroup$
    – ZenoCosini
    Jul 14, 2015 at 22:21

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