TITLE: Find the interval of convergence of the series
QUESTION [0 upvotes]: Assuming that $x$ is fixed, I wrote:

And then I used the Ratio test:

So I get the interval of $1$ $<$ $x$ $<$ $3$. However, my question is: how can $sin$ be convergent at all? I mean, its' values just randomly jump inside interval of $[$ $-1$$;$$1$$]$, how can it get some exact value?

REPLY [0 votes]: The way these tests are written, it's probably 
$  \sum \left( \sin\frac{\sqrt{n}}{n^2 + 1} \right) (x-2)^n$
which is much easier.
If it is, indeed,
$  \sum  \sin \left(\frac{\sqrt{n}}{n^2 + 1}  (x-2)^n \right)$
then it's a sine of a positive argument when $x > 2$ (and therefore, basically, your calculation works) and alternating series when $x < 2$. Thus, for $x < 2$, you can use alternating series test. 
Also, keep in mind the inequality $\sin |x| < |x|$.
How can $sin$ converge at all? If it's a sine of something small, then it does not randomly jump all over $[-1,1]$, but is also small.
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