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

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

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.

• Hint: Your series start with $n=1$, not with $n = 0$. – Martin R May 1 '15 at 10:29

As Martin R noticed, observe that you have $$\sum_{n=0}^\infty r^n=\frac1{1-r}, \quad |r|<1,$$ but
$$\sum_{n=\color{red}{1}}^\infty r^n=\frac {\color{red}{r}}{1-r}, \quad |r|<1.$$
• I just follow the formula here regentsprep.org/regents/math/algtrig/atp2/geoseq.htm to calculate the sum of geometric series so I don't get what the mistake I made as your explaination. Can you explain further? It say that the geometric series starting with n = 1 have the sum $\frac1{1-r}$ – aukxn May 1 '15 at 10:39
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.