Convergence of $\sum_{i=1}^\infty 9^{-n+2}4^{n+1}$

Question

Am I correct so far in finding if following sequence is convergent or divergent?

$\sum_{i=1}^\infty 9^{-n+2}4^{n+1}$

$9^{2}\times4\sum_{i=1}^\infty 9^{-n}*4^{n}$

$9^{2}\times4\sum_{i=1}^\infty 4^{n}/9^{n}$

Answer 1

This is the correct procedure. Once you have this step, you can see that this is a geometric series of common ratio $\frac{4}{9}<1\implies$ the series is convergent.

We can check the common ratio, because:

$$9^2*4\sum_{i=1}^{\infty}\frac{4^n}{9^n}=324\sum_{i=1}^{\infty}\big(\frac{4}{9}\big)^n=324\big(\frac{4}{9}+\frac{16}{81}+\cdots+\big)$$

And clearly, the ratio of terms, as seen in the closed form formula for the series is $\frac{4}{9}<1$.

Answer 2

Your procedure is correct the series is convergent so for finding sum upto infinity we have a formula $\frac {a}{1-r}$ so here first term is $4/9$ and also common ratio is $4/9$ so sum is $324.4/5$=$1296/5$.