Calculate the sum of $\sum_{n=1}^{+\infty} \frac{1+2^n}{3^n}$

Question

$$\sum_{n=1}^{+\infty} \frac{1+2^n}{3^n} = \sum_{n=1}^{+\infty} \frac{1}{3^n} + \sum_{n=1}^{+\infty} \frac{2^n}{3^n}$$

Each term is geometric series with $-1<r<1$ so they are all covergent. As the $\lim_{n\to+\infty}r^n=0$ for $-1<r<1$

$$\sum_{n=1}^{+\infty} \frac{1}{3^n} = \lim_{n\to+\infty}\frac{1-(\frac13)^n}{1-\frac13}=\lim_{n\to+\infty}\frac32( 1-(\frac13)^n) = \frac32$$

$$\sum_{n=1}^{+\infty} \frac{2}{3^n} = \lim_{n\to+\infty}\frac{1-(\frac23)^n}{1-\frac23}=\lim_{n\to+\infty}3( 1-(\frac23)^n) = 3$$

$$\to\sum_{n=1}^{+\infty} \frac{1}{3^n} + \sum_{n=1}^{+\infty} \frac{2^n}{3^n} = 3+\frac32 = \frac92=4.5$$

However when I use wolfram alpha to verify the result, the answer is $2.5$ means $\frac52$ not $\frac92$. I cant understand why it's like that.

Answer 1

As Martin R noticed, observe that you have $$ \sum_{n=0}^\infty r^n=\frac1{1-r}, \quad |r|<1, $$ but

$$ \sum_{n=\color{red}{1}}^\infty r^n=\frac {\color{red}{r}}{1-r}, \quad |r|<1. $$

Answer 2

You have difference two with the original solution, and this grows from this fact that $$ \sum_{k=1}^\infty r^n=\frac1{r-1}-1 $$ but you made the mistake $$ \sum_{k=1}^\infty r^n=\frac1{r-1} $$ twice. Which gives an error of value two.