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SUMMARY

The infinite series sum of \(\sum_{n=0}^{+\infty} \frac{\pi \cos(n)}{5^n}\) cannot be evaluated using the geometric series formula \(S = \frac{a_1}{1 - r}\) because it is not a geometric series due to the presence of \(\cos(n)\). Instead, the cosine function can be transformed into exponential form using the identity \(\cos(n) = \frac{(e^{i})^n + (e^{-i})^n}{2}\). This transformation allows the series to be expressed as a sum of geometric series, facilitating its evaluation.

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waealu
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Homework Statement


(Sorry, but I haven't mastered using the sigma notation in these forums yet).

Find the sum of the following infinite series: (n=0)^(inf) SIGMA ((pi)cos(n))/(5^n).


Homework Equations


I tried using the formula S=(a1)/(1-r).

I know that a=pi, but I can't find "r." The equation doesn't have a consistent rate. Should I be using a different method?

Thanks.
 
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Hi waealu!

The sum you have wrote down is not a geometric series, so the formula you mentioned is not applicable (yet).

The problem is the cos(n), that spoils the fun. But there is a way to change the cos(n) into exponents by using the formula

[tex]\cos(n)=\frac{(e^i)^n+(e^{-i})^n}{2}[/tex]

With this equation, you can express your series as a (sum of) geometric series.

Note, to display the sum with LaTeX, type

[ tex ] \sum_{n=0}^{+\infty} \frac{\pi \cos(n)}{5^n} [ /tex ]

without the spaces in the [ tex ] tags.
 

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