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Find the sum of the series or show that the series is divergent.

$$\sum_{n=0}^\infty \frac{5^n-2}{7^n}$$

So, I've established that this series is convergent via the comparison method; however, I'm unable to find the sum. I can't seem to find the ratio, so I'm assuming it's a non-geometric series. I tried finding a pattern by expanding the series, but have not come to any conclusions. Below you can see my expansion.

$$\sum_{n=0}^\infty (\frac{5}{7})^n - (\frac{2}{7^n}) = (-1) + (\frac{5}{7} - \frac{2}{7}) + (\frac{25}{49} - \frac{2}{49}) + (\frac{125}{343} - \frac{2}{343}) + ...$$

This is the beginning of my calc 2 course, so the scope of my knowledge is limited. Thank you for taking the time to read this problem. I appreciate any and all help.

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I suppose that you know $\sum \left(\frac{5}{7}\right)^n$ and $\sum \frac{2}{7^n}$ converge. Then $$ \sum_{n=0}^{\infty} \frac{5^n -2}{7^n} =\sum_{n=0}^{\infty} \left(\frac{5}{7}\right)^n - \sum_{n=0}^{\infty}\frac{2}{7^n}. $$ Can you proceed from this?

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  • $\begingroup$ Thanks for helping out choco_addicted! Actually, seeing it written out like that clears everything up! $\endgroup$ – Joey J May 20 '16 at 7:15
  • $\begingroup$ Got $$\frac{7}{6}$$ $\endgroup$ – Joey J May 20 '16 at 7:21
  • $\begingroup$ @JoeyJ That is a correct answer. Wolframalpha here $\endgroup$ – choco_addicted May 20 '16 at 7:23
  • $\begingroup$ awesome! Hope your night is going well :) $\endgroup$ – Joey J May 20 '16 at 7:51
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Hint:

If $|r|<1$, then we have $$\sum_{n=0}^\infty r^n=\frac{1}{1-r}$$

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