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I'm having trouble finding the radius of convergence for the series $ \sum_{n=1}^{\infty} \frac{(-1)^{n}.z^{n(n+1)}}{n}$, because I can't find its $n$-th coefficient. My aim is to use the theorem that says:

Theorem:

For a given power series $ \sum_{n=1}^{\infty} a_n(z-a)^n$ define the number $R, 0 \leq R \leq \infty$ by: $\frac{1}{R} = \lim \sup |a_n|^{1/n}$

Then:

(a) if $|z-a|<R$, the series converges absolutely;

b) if $|z-a|>R$, the terms of the series become unbounded and so the series diverges;

(c) if $0 \leq r \leq R$, then the series converges uniformly on ${z:|z|\leq r}$

Moreover, the number $R$ is the only number having properties (a) and (b).

But how to find $a_n$ in a series like that?

Defining a function for the exponent, $f(n) = n.(n+1)$, I observed that:

$(f(i))_{i=1}^{\infty} = (2,6,12,20,30,42,56,...,n(n+1),...)$

Thanks.

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    $\begingroup$ Surely you can find some $R$ such that if $|z|>R$ then "the terms of the series become unbounded" while if $|z|<R$ then "the series converges absolutely"! Note that the theorem you recall does not require to identify $a_n$... $\endgroup$ – Did Mar 10 '16 at 21:48
  • $\begingroup$ Why the . for multiplication? $\endgroup$ – zhw. Mar 11 '16 at 22:06
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The trick when your $z$ is indexed weird is to change the $\frac{1}{n}$ in the formula to match the indexing on the $z$, so look at $\limsup(\frac{1}{n})^{\frac{1}{n(n+1)}}$.

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  • $\begingroup$ But why $\frac{1}{n}$ ? Is it the n-th coefficient? I can't see.. $\endgroup$ – user286485 Mar 11 '16 at 13:35
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    $\begingroup$ Yea, disregarding the mismatched indices the coefficients are of the form $\frac{(-1)^n}{n}$, but we are taking the absolute value so we disregard the $(-1)^n$ $\endgroup$ – siegehalver Mar 11 '16 at 15:48
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If $|z|<1,$ the series converges absolutely. If $|z|>1,$ the series diverges. That alone is enough to tell you the radius of convergence is $1.$

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