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im looking for hints on how to do: $\sum\limits_{n=4}^{n= \infty } \frac{2^n + 8^n}{10^n} = ?$

I thought this may have had something to do with geometric series but nothing obvious comes up

thanks in advance

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    $\begingroup$ Why not just write it as a sum of two geometric series? $\endgroup$ – orion Jun 13 '15 at 10:41
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Hint: you can write $\sum\limits_{n=4}^{n= \infty } \frac{2^n + 8^n}{10^n} = \sum\limits_{n=4}^{n= \infty } \frac{2^n }{10^n} + \sum\limits_{n=4}^{n= \infty } \frac{8^n }{10^n} $ because the two series on the right hand side converge. And then you have two geometric series, which you can probably solve.

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$$\sum\limits_{n=4}^{\infty } \frac{2^n + 8^n}{10^n} \\ = \sum\limits_{n=4}^{\infty }\left( \frac{1}{5^n}+\frac{4^n}{5^n}\right)\\=\frac{1}{1-\frac{1}{5}}-1-\frac{1}{5}-\frac{1}{25}-\frac{1}{125}+\frac{1}{1-\frac{4}{5}}-1-\frac{4}{5}-\frac{16}{25}-\frac{64}{125}\\=0.002+2.048\\=2.05$$

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\begin{align*} \sum\limits_{n=4}^{\infty} \frac{2^n + 8^n}{10^n} &= \sum\limits_{n=4}^{\infty } \frac{2^n}{10^n}+\sum\limits_{n=4}^{\infty } \frac{8^n}{10^n} \\ &=\sum\limits_{n=4}^{\infty } \left(\frac{1}{5}\right)^n+\sum\limits_{n=4}^{\infty } \left(\frac{4}{5}\right)^n \\ &=-1-\frac{1}{5}-\frac{1}{25}-\frac{1}{125}+\sum\limits_{n=0}^{\infty } \left(\frac{1}{5}\right)^n-1-\frac{4}{5}-\frac{16}{25}-\frac{64}{125}+\sum\limits_{n=0}^{\infty } \left(\frac{4}{5}\right)^n \\ &=-1-\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-1-\frac{4}{5}-\frac{16}{25}-\frac{64}{125}+\frac{1}{1-\frac{1}{5}}+\frac{1}{1-\frac{4}{5}} \\ &= -3-\frac{1}{5}-\frac{17}{25}-\frac{65}{125}+\frac{5}{4}+5 = 2.05 \end{align*}

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