Does this infinite geometric series diverge or converge?

Question

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.

Answer 1

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 $$\sum_{n=2}^\infty ar^n=\frac{ar^2}{1-r}.$$

Answer 2

The formula you used is only suitable when $|r| < 1$. If that isn't enough, consider that a series only converges if $a_n \rightarrow 0$ as $n \rightarrow \infty$. If $|r| > 1$ then $a_n$ does not go to 0.