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.

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    $\begingroup$ Is it not even a well defined series. The question of convergence arises only if the first term is omitted. $\endgroup$ 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
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    $\begingroup$ 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. $\endgroup$ Mar 4, 2019 at 11:50

1 Answer 1


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 $=2.$


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