0
$\begingroup$

Given the power series $$\sum_{n=0}^\infty\frac{(-1)^n}{2^n\sqrt{n}}(x-1)^n,$$ I found using the ratio test that the radius of convergence is $R=1/(1/2)=2$, but I'm a little confused as to why this infinite sum even exists since the first term, for $n=0$, involves dividing by zero. So from a naive point of view, it would seem that the infinite sum should not converge. Does it actually converge?

Mathematica tells me $R=2$ also, but it also complains about division by zero if I try to sum to infinity. This could be a limit of Mathematica, however.

It could be the sum does converge even with the division by zero since the sum is infinite, and as a Taylor series represents some perfectly finite analytic function. Furthermore, by analogy, I think there are certain integrands which diverge at $a$, say, yet whose definite integrals converge over $[a,b]$, say, although I can't remember an explicit example right now.

$\endgroup$
5
  • 2
    $\begingroup$ Is it not even a well defined series. The question of convergence arises only if the first term is omitted. $\endgroup$
    – Kavi Rama Murthy
    Mar 4, 2019 at 11:40
  • $\begingroup$ indeed, this power series is not defined as the $0$-th term is not defined, in particular you need to either get rid of it or force it to be special term. however, I assume you just want to kill that one off. $\endgroup$
    – Enkidu
    Mar 4, 2019 at 11:41
  • 2
    $\begingroup$ Convergence only concerns the tail of the series, not any particular term. Your series should probably start with $n=1$ since it isn’t well-defined otherwise. The individual terms must certainly exist before you can even begin to consider questions of convergence. $\endgroup$
    – MPW
    Mar 4, 2019 at 11:42
  • $\begingroup$ @KaviRamaMurthy So as it stands, including the $n=0$ term, to ask what its radius of convergence $R$ is would be meaningless? $\endgroup$
    – gone
    Mar 4, 2019 at 11:48
  • $\begingroup$ @Antinous Yes. The question does not make sense. $\endgroup$
    – Kavi Rama Murthy
    Mar 4, 2019 at 11:50

1 Answer 1

Reset to default
0
$\begingroup$

I think that $\sum_{n=0}^\infty\frac{(-1)^n}{2^n\sqrt{n}}(x-1)^n$ is a typo and it should read $\sum_{n=1}^\infty\frac{(-1)^n}{2^n\sqrt{n}}(x-1)^n$.

The radius of convergence of $\sum_{n=1}^\infty\frac{(-1)^n}{2^n\sqrt{n}}(x-1)^n$ is indeed $=2.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.