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

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.

  • 3
    $\begingroup$ Hint: Your series start with $n=1$, not with $n = 0$. $\endgroup$ – 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

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

  • $\begingroup$ 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 $\frac1{1-r}$ $\endgroup$ – aukxn May 1 '15 at 10:39
  • $\begingroup$ @aukxn If you want to "follow the formula there", at least do not forget to include the first term as a factor... $\endgroup$ – Did May 1 '15 at 10:45
  • $\begingroup$ oh, I got it, thank you so much $\endgroup$ – aukxn May 1 '15 at 10:47

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.


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