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I was shown this example, but do not understand the procedure for doing this and why it's correct.

Show $\sum x^{n!}$ converges for $x \in (0,1)$.

Way it was shown: Write this as $x+x^2+0x^3+0x^4+\dots$.i.e. $\sum c_kx^k$ where $c_k$ is $1$ if $k=n!$ for some $n$, and $0$ otherwise. Now use the comparison:$|c_kx^k| \leq x^k$ so compare with geometric series.Can someone explain what is going on here with this method? Why can you write the series like this?Thanks

Scott Frazier
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    Do you agree or not that $\sum_{n=1}^\infty x^{n!}=\sum_{k=1}^\infty c_kx^k$, with $c_k=1$ if $k=n!$ for some $n\in\Bbb N$ and $0$ otherwise? – José Carlos Santos Feb 09 '21 at 22:41
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    @JoséCarlosSantos ok now I think I understand it, thanks. So not all the terms past $x$ and $x^2$ are zero, that was my belief, and I think that is what was confusing me.And I must admit a pretty absurd belief given the definition of the coefficients $c_k$ – Scott Frazier Feb 09 '21 at 22:46
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    Indeed,$$\sum_{k=1}^\infty c_kx^k=x+x^2+0\times x^3+0\times x^4+0\times x^5+x^6+0\times x^7+\cdots+0\times x^{23}+x^{24}+0\times x^{25}+\cdots$$ – José Carlos Santos Feb 09 '21 at 22:54

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This response is somewhat off-topic because the OP specifically asked for a proof that involves the comparison test. However, comments that preceded this answer have provided such a proof. Therefore, I regard it as open season to provide an alternative approach.

If you have a strictly increasing sequence of numbers, that are bounded above, then the sequence must be convergent.

You are given that $\sum_{k=0}^{\infty} x^k$ is a convergent series. Denote its limit as $L$.

Define the sequence $a_0, a_1, a_2, \cdots,$ by $a_k = \sum_{i=0}^k x^{(i!)}.$
Then the problem reduces to showing that $\langle a_k\rangle$ is a convergent sequence.

You know that this sequence is strictly increasing.

As other responses have indicated, the sequence is bounded above by $L$.

user2661923
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