Does this infinite geometric series diverge or converge? I am given the infinite geometric series $ \sum_{n=2}^{\infty}\left(5^{n-1}\right)\left(\frac{9}{10}\right)^n $ and told that it diverges. However, my calculations indicate otherwise. 
My calculations are as follows.
$ \sum_{n=2}^{\infty}\left(5^{n-1}\right)\left(\frac{9}{10}\right)^n $
$ = \sum_{n=2}^{\infty}\left(5^{-1}\right)\left(5^{n}\right)\left(\frac{9}{10}\right)^n $
$ = \sum_{n=2}^{\infty}\left(5^{-1}\right)\left(\frac{9}{2}\right)^n $
$ \dfrac{a}{1-r} $ is the formula for the sum of an infinite geometric series.
$ \therefore \dfrac{a}{1-r} = \dfrac{5^{-1}}{1 - \frac{9}{2}}$
$ = \dfrac{\frac{1}{5}}{\frac{-7}{2}} $
$ = \dfrac{-2}{35} $
Therefore, the infinite geometric series converges to $ \dfrac{-2}{35} $.
Please check my reasoning and specify where I am incorrect, why I am incorrect, and the correct reasoning to arrive at the correct answer.
Thank you.

If we apply your reasoning, 
$$
\sum_{n=1}^\infty 2^n=\frac{2}{1-2}=-2.
$$
You should ask yourself how you get a negative result by adding all positive terms. 
The reason is that the formula for the geometric series $\sum_n r^n$ applies when the series is convergent, which requires $|r|<1$. 
On another note, the formula for your series (had it been convergent) would have been
$$\sum_{n=2}^\infty ar^n=\frac{ar^2}{1-r}.$$
