I am trying to find the sum of the following series:

$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}}$$

which I rewrote as

$$\sum_{n=1}^{\infty} \left(\frac{1}{9^n}+ \left(\frac{7}{9}\right)^n\right)$$

I am assuming that it is a geometric series and the initial value is

$$a_1=\frac{1}{9} + \frac{7}{9}$$

I also see that

$$a_2 = \frac{1}{9^2} + \frac{7^2}{9^2}$$

I know that in a geometric series the first term is $a$ and the second term is $ar$.

This allows me to see that


which when solved for $r$ gives the value $\frac{25}{36}$.

Using the formula to find the sum of a geometric series $\frac{a}{1-r}$, I find that the sum is equal to $\frac{32}{11}$.

But this value is incorrect and the sum is actually $\frac{29}{8}$. How does one find that value?

  • 1
    $\begingroup$ It is not a geometric series. But you can express it as the sum of two geometric series: $\Sigma_n 1/9^n$ and $\Sigma_n 7^n/9^n.$ $\endgroup$ – mfl Nov 5 '14 at 0:33
  • $\begingroup$ Remember that $a^n+b^n\neq(a+b)^n$, I think that was what confused you. $\endgroup$ – David Nov 5 '14 at 6:59

This is not a geometric series.

But it is the sum of two geometric series:

$$\sum_{n=1}^{\infty} {\frac{1+7^n}{9^n}} = \sum_{n=1}^{\infty} {\frac{1}{9^n}}+ \sum_{n=1}^{\infty} {\frac{7^n}{9^n}} =\frac 19 \frac 1{1-\frac 19} +\frac 79\frac 1{1-\frac 79}$$

because $$ \left|\frac 19\right|<1 \\\left|\frac 79\right|<1 $$

  • 2
    $\begingroup$ The sums start at $n=1$ not $n=0$ making the formula for geometric series different. $\endgroup$ – coffeemath Nov 5 '14 at 0:35
  • $\begingroup$ @coffeemath oh right, thanks! $\endgroup$ – mookid Nov 5 '14 at 0:36
  • $\begingroup$ Beat me too it. $\endgroup$ – Dair Nov 5 '14 at 0:38

Hint: Try re-doing it as two separate series, $\sum(1/9)^n$ and $\sum(7/9)^n.$ The first has first term $1/9$ and common ratio $1/9$ while the second has first term $7/9$ with common ratio $7/9.$


$$ \frac{\frac19}{1-\frac19}+\frac{\frac79}{1-\frac79}=\frac18+\frac72=\frac{29}{8}. $$

  • $\begingroup$ is this really helpful? $\endgroup$ – mookid Nov 5 '14 at 0:36
  • $\begingroup$ @mookid Something must be left to OP's own work... $\endgroup$ – Przemysław Scherwentke Nov 5 '14 at 0:39
  • 3
    $\begingroup$ I don't think this is the tricky part. $\endgroup$ – mookid Nov 5 '14 at 0:40

\begin{align*}\sum_{n=1}^\infty\frac{1+7^n}{9^n} &= \sum_{n=1}^\infty\bigg[ \frac{1}{9^n} + \frac{7^n}{9^n}\bigg] \\ &= \sum_{n=1}^\infty\frac{1}{9}\bigg(\frac{1}{9}\bigg)^{n-1} + \sum_{n=1}^\infty \frac{7}{9}\bigg(\frac{7}{9}\bigg)^{n-1} \\ &=\frac{1}{9}\bigg(\frac{1}{1-\frac{1}{9}}\bigg)+\frac{7}{9}\bigg(\frac{1}{1-\frac{7}{9}}\bigg) \\ &= \frac{1}{8}+\frac{7}{2} \\ &= \frac{29}{8} \end{align*}


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