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Was reading Thomas Calculus and found this loophole which I can't fill up could someone give an outline of this prove/explain? Much appreciated!

Prove that SEQUENCE (not series) of terms of a power series must diverge outside the radius of convergence of a power series, except possibly ON the boundary point of interval of convergence

eg. consider the series sum(x^n/n) radius of convergence: [-1,1), sequence nevertheless converges at x=1 (harmonic series), but diverges elsewhere, we need to show that other series like as the harmonic series does not exist elsewhere outside the radius of convergence

What I tried was considering power series at c and c+epsilon, but the binomial expansion of the latter became quite complicated and I do not know how to use the fact that the series diverges for all numbers greater than c

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"sequence nevertheless converges at x=1 (harmonic series)", is false. The harmonic series is divergent. –  George V. Williams Jan 1 '13 at 23:57
    
@GeorgeV.Williams: the sequence the sum is over converges; not the sequence of partial sums. That is, $\lim x^n/n$ converges for all $|x|\leq 1$ while it diverges if $|x|>1$ (as l'Hopital's shows). –  Clayton Jan 2 '13 at 0:57
    
∑n=1∞xn is not a power series –  Zhanfeng Lim Jan 2 '13 at 1:29

3 Answers 3

The fact that power series have a radius of convergence is usually proved by the root test, which boils down to comparing it to a geometric series. Geometric series diverge when their sequence diverge and converge when their sequence converges, except when the ratio is 1. The ratio being 1 corresponds to the boundary for power series. So, I would take the proof that power series have a radius of convergence, and tweak the part that uses the root test.

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Thank you! That was concise and insightful –  Zhanfeng Lim Jan 2 '13 at 1:30

I assume that the radius of convergence $R<\infty$, otherwise there is nothing to do.

You have $\frac{1}{R} = \limsup_{n \to \infty} \sqrt[n]{|a_n|}$. Hence for some subsequence, you have $\frac{1}{R} = \lim_{k \to \infty} \sqrt[n_k]{|a_{n_k}|}$, or equivalently, $1 = \lim_{k \to \infty} \sqrt[n_k]{|a_{n_k}|R^{n_k}}$.

Now suppose that $|x-x_0| > R$, and consider the $n_k$th term of the power series. You have $|a_{n_k}||x-x_0|^{n_k} = |a_{n_k}R^{n_k}| (\frac{|x-x_0|}{R})^{n_k} = (\sqrt[n_k]{|a_{n_k}|R^{n_k}}\frac{|x-x_0|}{R})^{n_k}$. Since $\lim_{k \to \infty} (\sqrt[n_k]{|a_{n_k}|R^{n_k}}\frac{|x-x_0|}{R})^{n_k} = \infty$, the terms $a_{n_k}(x-x_0)^{n_k}$ do not converge to zero, hence the series diverges.

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Let $\sum_{n=1}^\infty a_n (x-a)^n$ be your power series.

Then, the radius of convergence is given by

$$r= \frac{1}{\limsup_n \sqrt[n]{a_n}} \,.$$

then if $|x-a| >r$ we have

$$\limsup_n |x-a|\sqrt[n]{a_n}=b >1 \,.$$

This implies that for infinitely many $n$ we have

$$\left|a_n (x-a)^n \right| > (\frac{b+1}{2})^n $$

and since $(b+1)>2$ youa re done.

Alternately

Assume by contradiction that $\lim_n a_n(x-a)^n =l$ for some $|x-a|>r$. Then $a_n(x-a)^n$ is bounded. Pick some $y$ so that

$$r< |y-a| < |x-a|$$

Then

$$a_n |y-a|^n = a_n |x-a|^n (\frac{|y-a|}{|x-a|})^n \leq b (\frac{|y-a|}{|x-a|})^n$$

This shows that $\sum a_n |y-a|^n$ is convergent since it is less than a convergent geometric series. But this contradicts the fact taht $|y-a| >r$.

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Thanks! Your alternate solution was clear to me. I haven't had a course in analysis though so I need some time for the first solution –  Zhanfeng Lim Jan 2 '13 at 1:35

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