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The infinite geometric series $\sum_{n=0}^\infty x^n=\frac{1}{1-x}$ and so is convergent for $|x|<1$.

While working on a trigonometry problem I came across the case of what happens when $x=0$. Clearly the sum is $1$ with the convention $0^0=1$ (this convention is discussed in several threads here on MathSE).

However Wolfram Alpha says about $\sum_{n=0}^\infty 0^n$:

(sum does not exist due to singularity at one or more evaluation points)

By the geometric series test, the series diverges.

My assumption is that Wolfram Alpha doesn't know what to do with $0^0$. There have also been several threads on MathSE about various ways in which Wolfram Alpha fails or uses certain unstated conventions.

Is there something else I am missing?

Ultimately I am interested in $\sum_{n=0}^\infty (\arctan(\sin(\theta)))^n$ (which is not hard to analyze), and had a problem evaluating it at $\theta=0$ on Wolfram Alpha. I'm including this problem on an upcoming test, so I wan't to make sure I wasn't overlooking anything. I suppose I could just start at $n=1$ to avoid any possible issue.

EDIT: The answer was provided by @Masacroso in the first comment below. See indeterminate forms on Wolfram Alpha. Wolfram Alpha does indeed consider $0^0$ to be undefined.

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jdods
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    wolframalpha doesnt use the convention $0^0:=1$, see here. Also $(\arctan(\sin 0))^0=0^0$ so this is the same problem for wolframalpha. – Masacroso Feb 15 '18 at 16:16
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    With the convention $0^0=1$ the series is indeed $1$ as you say. This is just a difference in convention with wolfram alpha. – Dave Feb 15 '18 at 16:17
  • Ok, thanks for confirming my suspicions. @Masacroso, your link is the solution to my question. Thanks. – jdods Feb 15 '18 at 16:20
  • The nice thing about Mathematica is that it allows you to override some of its conventions. One example is $\mathtt{Unprotect[Power]; 0^0=1}$. – Somos Feb 15 '18 at 19:47
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    I just discovered that Octave (Matlab) gives $0^0=1$ as well. Good to be aware of! – jdods Feb 15 '18 at 20:25

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