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I am doing Apostol sections concerning series/sequences. In all his statements of tests, like ratio and root tests, it is mentioned that the tested series are non-negative, which automatically precludes the use of these tests for the complex series, since complex numbers cannot be non-negative. However, in Wiki, it is mentioned that the root and ratio tests are actually used with the complex series too. Moreover, some of Apostol exercises are not easily solvable without using these tests with the complex series.

But how does it work? Both the root and ratio tests are based on the comparison with the geometric series. In order to use the root test, we first need to take the absolute value of the complex sequence, because the limit has to be compared to 1, which is a real number. Thus, even if the root test shows that the convergence, it only shows the absolute convergence! It says nothing about conditional convergence.

An example from Apostol:

$\sum\limits_{n = 1}^\infty (1 + \frac{1}{n})^{n^2} z^n, z \in \mathcal{C}$

By the root test we get:

$|a_n|^{\frac{1}{n}} = (1 + \frac{1}{n})^n |z| = e|z|$ as $n \rightarrow \infty$

Now, we can see for which values of $z$ the series converges. If $|z| < 1/e$, then the series converges absolutely. If $|z| > 1/e$, the series diverges, if $|z| = 1$, the result is inconclusive.

By saying the series diverges in the case $|z| > 1/e$, why? To me, it says that the series does not converge absolutely (absolutely diverges?), however, it does not say if the series can still converge conditionally if $z > 1/e$.

Moreover, is it valid to test the above series without the modulus operation? So that $a^{\frac{1}{n}} = ez, n \rightarrow \infty$. What does it tell us then, and how is it compared to 1, since complex numbers are unordered?

Can someone clarify?

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    $\begingroup$ Taking the modulus is an essential part of the test, just like the absolute value is needed in the root test for real series. $\endgroup$ – RRL Jan 3 at 23:53
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Recall the term test. For any power series (real or complex) $\sum b_n z^n$, a necessary condition for convergence is $\lim_{n \to \infty} b_nz^n = 0$. If $b_nz^n \not\to 0$ as $n \to \infty$ then the power series must diverge.

Suppose in applying the root test we find $\lim_{n \to \infty}|b_n|^{1/n} = \alpha$. (This just generalizes your specific example where $b_n = (1 + 1/n)^{n^2}$ and $\alpha = e$.)

If $|z| > 1/\alpha$ we then have $\lim_{n \to \infty}|b_n|^{1/n}|z|= \alpha|z| > 1$. By definition of a limit, given any $r$ with $\alpha|z| > r> 1$, there exists $N \in \mathbb{N}$ such that $|b_n|^{1/n}|z| > r$ for all $n > N$.

This implies that for all $n > N$ we have $(|b_n|^{1/n}|z|)^n = |b_nz^n| > r^n > 1$ and

$$\lim_{n \to \infty}|b_nz^n| > \lim_{n\to \infty} r^n = +\infty$$

This shows that the sequence of complex numbers $b_nz^n$ cannot converge to $0$, otherwise, both real and imaginary parts and, hence, the modulus $|b_nz^n|$ must converge to $0$. By the term test the series $\sum b_nz^n$ must diverge if $|z| > 1/\alpha$.

So we see that the root test does not merely "say that the series does not converge absolutely" when $|z| > 1/e$ in your example. It cannot converge conditionally in this case.

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  • $\begingroup$ So, if I understood right, the root test equates the conditional and absolute convergences? So, if the root test limit value $\alpha |z| \neq 1$, then anytime the series converges absolutely, it will converge conditionally, and every time it converges conditionally, it will converge absolutely too? $\endgroup$ – John Jan 4 at 0:02
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    $\begingroup$ Yes. I would reserve the term conditional though when a series converges but not absolutely. in this case it converges both absolutely and non-absolutely. $\endgroup$ – RRL Jan 4 at 0:15
  • $\begingroup$ great explanation, thx! :) $\endgroup$ – John Jan 4 at 0:20

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