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I am wondering if my proof is valid.

Suppose that the sequence $(a_n)$ is bounded but that the series $\sum{a_n}$ diverges. Prove that the radius of convergence of the power series $\sum{a_n}x^n$ is equal to $1$

My attempt:

The sequence is bounded so, $\limsup|a_n|^{1/n}=1.$ Let this $=\alpha$

Let $b_n=a_nx^n$ and then use the root test.

$\beta =\limsup|b^n|^{1/n}=\limsup|a_nx^n|^{1/n}=|x|\limsup|a_n|^{1/n}=|x|\alpha$

Now, we know that $\sum{a_n}x^n$ converges when $|x|\alpha \lt 1$

But since, $\alpha=1$, then the radius of convergence is thus $1$

Is this valid? Could it be improved?

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    $\begingroup$ The conclusion that $\limsup |a_n|^{1/n}=1$ cannot follow only from $a_n$ being bounded. The condition that $\sum a_n$ diverges should be used. A good part of the substance of the proof lies in proving that first claim it better be shown. $\endgroup$
    – Marja
    Aug 10, 2017 at 0:41

2 Answers 2

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Your proof is almost valid, but unfortunately

The sequence is bounded so, $\limsup|a_n|^{1/n}=1.$ Let this $=\alpha$

is wrong for several reasons. First, it wasn't clear what $\alpha$ is; I had to read almost to the end to guess that you mean $\alpha := \limsup |a_n|^{1/n}$ (but that's just a writing issue). More importantly, the logic is not correct. The fact that $a_n$ is bounded does not imply this - for example, $a_n = 2^{-n}$ wouldn't have it. You need to also invoke the fact that the series $\sum_n a_n$ diverges.

To fix this: We have $\alpha \le 1$ because the sequence is bounded. If $\alpha < 1$, then root test would imply convergence of $\sum_n a_n$, so by contraposition we find that $\alpha \ge 1$. Hence $\alpha = 1$ and the rest of your proof is fine.

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  • $\begingroup$ @TBongers, can you please justify this statement: "We have $\alpha \le 1$ because the sequence is bounded." $\endgroup$
    – makansij
    Jun 7, 2019 at 23:09
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    $\begingroup$ @nundo $a_n$ is bounded, so $\vert a_n\vert \leqslant M$, then $\vert a_n \vert \leqslant M^{1/n}$. By taking superior limits, $\varlimsup \vert a_n \vert ^{1/n} \leqslant \lim M^{1/n} = 1$. $\endgroup$
    – xbh
    Jun 8, 2019 at 0:07
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If we are allowed to say without proof that since $a_k$ is bounded, then $\limsup_{k \to \infty} \sqrt[k]{a_k} \le 1$, then this implies by Cauchy Hadamard, $R\le 1$. Then, since $\sum_{k=0}^{\infty} {a_k} $ diverges, then if $|x| > 1$ then $\sum_{k=0}^{\infty} {a_k} x^k $ diverges. That implies that $R$ cannot be $> 1$, so $R\le 1$. Since $R \le 1$ and also $R\ge 1$, then $R=1$.

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