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Let $\{ a_{n} \}$ be a decreasing sequence such that $\lim_{n \rightarrow \infty} a_{n} =0$. Prove that the power series $\sum_{n=0}^{\infty} a_{n} x^{n}$ converges if $|x| \leq 1$ and $ x \neq 1$.

I was trying to see if I could use the ratio test . But while its clear that $ \limsup_{n \rightarrow \infty} |\frac{a_{n+1}}{a_{n}}| <1$ it is not clear to me that this sequence or any subsequence of it is convergent(as I can't tell if the sequence is monotone though it's clearly bounded). Yet, if it is equal to 1, then I believe the ratio test applies and then we can maybe use the alternating series test to show that the series converges for $x = -1$ as well. Thanks

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It's not at all clear that $\limsup_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n} \right| < 1$. Try e.g. $a_n = 1/n$. –  Robert Israel Jan 1 '13 at 2:52
    
ah right, thanks –  Jmaff Jan 1 '13 at 2:56

3 Answers 3

up vote 1 down vote accepted

The comparison test proves convergence for $|x|<1$, since for sufficiently large $n$, $|a_n|\leq 1$.

For $|x|=1$, and $x\neq 1$, we can use summation by parts. If $|x|=1$, then $x=e^{i \theta}$, and so the partial sum is $$\sum_{n=1}^N a_n e^{i n\theta}.$$ From here, we can use Dirichlet's test, which is proven by summation by parts. This shows that the series converges for $0<\theta<2\pi$.

Edit: What I said previously regarding the alternating series test was incorrect. Thanks to sos440 for pointing out my mistake.

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Alternating series test is not adequate for non-real $x$ with $|x| = 1$. Dirichlet test directly works for this case. –  sos440 Jan 1 '13 at 3:13
    
@sos440: I updated the answer. –  Eric Naslund Jan 1 '13 at 3:30
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As far as I know, alternating series test requires monotonicity of the magnitude of the sequence. But this is in general not true for $(b_n)$. For example, if we let $a_n = n^{-1}$ and $\theta = 1$, then numerical test shows that $|b_n|$ oscillates though it has a decreasing envelope. Of course a much detailed analysis may allow us to overcome this impediment, I guess that the trade-off will be tremendous. That's why I think alternating series test is inadequate in this situation. –  sos440 Jan 1 '13 at 3:42
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@LevLivnev, For $\theta$ away from $2\pi \Bbb{Z}$, we have $$ \sum_{n=1}^{N} e^{in\theta} = \frac{e^{i\theta}\left( e^{iN\theta} - 1 \right)}{e^{i\theta} - 1}. $$ Then the rest is clear. –  sos440 Jan 1 '13 at 4:35
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@sos440 Beautiful, it's perfectly clear now. –  user54147 Jan 1 '13 at 5:48

Hint: use a comparison test for $|x| < 1$. For $x = -1$ you can use the alternating series test. For complex $x$ with $|x|=1$, use Dirichlet's test.

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If $|x|<1$, then we have $$\sum_{n=0}^\infty a_nx^n\leq\sum_{n=0}^\infty a_0x^n=\frac{a_0}{1-x},$$ by geometric series. If $|x|=1$, this means $x=-1$, and the alternating series test proves convergence.

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The inequality is often false if $x<0$. –  Jonas Meyer May 29 '13 at 1:52

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