Why wolfram alpha claimed that this $\sum_{n=1}^{\infty}\sin (\frac{n}{\sqrt{n!}})$ is converge by test and in the same time is diverge?

I'm confused why wolfram alpha claimed that this sum $$\sum_{n=1}^{\infty}\sin \left(\frac{n}{\sqrt{n!}}\right)$$ is convergent by test criterion, and in the same time is divergent in result below in the picture?

In my guess it probably shows us the obscurity of evaluation of that series, or something like that or convergence test in Wolfram alpha is not enough to show wether that series is diverge or converge?

• Seems to be a bug in Wolfram Alpha, since the sum definitely does converge. You can report it using the feedback button at the bottom of the page. – Nate Eldredge Mar 30 at 0:04
• Yes , i have tried now but i didn't succeed – zeraoulia rafik Mar 30 at 0:12

The backend has an error causing it to believe that $$\sum_{n=1}^k \sin\left( \frac{n}{\sqrt{n!}} \right)$$ is $$-\frac{i e^{-\frac{i k}{\sqrt{\text{Sum\grave{ }SumqBaseDump\grave{ }u\3851}!}}} \left(-1+e^{\frac{i k}{\sqrt{\text{Sum\grave{ }SumqBaseDump\grave{ }u\3851}!}}}\right) \left(-1+e^{\frac{i (k+1)}{\sqrt{\text{Sum\grave{ }SumqBaseDump\grave{ }u\3851}!}}}\right)}{2 \left(-1+e^{\frac{i}{\sqrt{\text{Sum\grave{ }SumqBaseDump\grave{ }u\3851}!}}}\right)}$$ where "$$\text{Sum\grave{ }SumqBaseDump\grave{ }u\3851}$$" is an internal symbol that should never have appeared in any result returned by Sum[].
The series converges, to about $$4.322187510593720884347337899899583088\dots$$ because $$\frac{n}{\sqrt{n!}}$$ approaches $$0$$ exponentially rapidly and sine of a very small positive number is very slightly less than that number. So this series is bounded by a geometric series and the comparison test shows it converges.