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$\sum_{n=1}^\infty z^{n!}$

Here is what I've got so far

Claim: The above series converges for $|z|<1$.

Pick $|z|. Then for all $n$, $|z^{n!}|<=r^{n!}$.

So $\sum\limits_{n=1}^\infty r^{n!}$ is a majorant for $\sum\limits_{n=1}^\infty z^{n!}$.

$\sum\limits_{n=1}^\infty r^{n!}$ is a real series so we can test for convergence.

This is where I get stuck, I've tried the ratio test but that doesn't seem to work and I can't think of a function that would work for the comparison test.

  • 0
    Aah cool so i can just use the comparison test on that. Thanks2012-11-04

4 Answers 4

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Try the root test, instead of the ratio test.

Alternately, observe that $\sum\limits_{n=0}^\infty r^{n!}$ is a subseries (in a sense) of the convergent geometric series $\sum\limits_{k=0}^\infty r^k=\frac1{1-r}$, so we can use comparison test that way. Hint: Write $\sum\limits_{n=0}^\infty r^{n!}=\sum\limits_{k=0}^\infty c_kr^k$, with the $c_k$'s defined appropriately.

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$\sum_{n=1}^\infty z^{n!}= \sum_{n=1}^\infty a_{k}z^{k}$ then $a_{k}=1$ for $k=n!$ and $a_{k}=0$ $k \neq n!$ then apply root test

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Since $|z|<1$, then $\frac{|z|^{(n+1)!}}{|z|^{n!}} = |z|^{n\cdot n!} \le |z| < 1$ for $n\ge1$, so the series is absolutely convergent by the ratio test.

  • 0
    No problem, oen.2012-11-04
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Forget this is an entire series and define $x_n=z^{n!}$ for every $n\geqslant1$.

  • If $|z|\geqslant1$, then $|x_n|\geqslant1$ for every $n$ hence the sequence $(x_n)_n$ does not converge to zero hence the series $\sum\limits_nx_n$ diverges.
  • If $|z|\lt1$, then $|x_n|\leqslant|z|^n$ for every $n$ (since $n!\geqslant n$) and the series $\sum\limits_n|z|^n$ converges hence the series $\sum\limits_nx_n$ converges (absolutely).