logistic_guy
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\(\displaystyle \textcolor{indigo}{\bold{Solve.}}\)
\(\displaystyle \sum_{k=0}^{\infty}\frac{5^k}{3}\)
\(\displaystyle \sum_{k=0}^{\infty}\frac{5^k}{3}\)
infinite series - 7
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logistic_guy
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Let us compute a few terms.
\(\displaystyle \sum_{k=0}^{\infty}\frac{5^k}{3} = \frac{1}{3} + \frac{5}{3} + \frac{5^2}{3} + \cdots\)
Well this is a geometric series.
\(\displaystyle \sum_{k=0}^{\infty}\frac{5^k}{3} = \lim_{k\rightarrow \infty}\frac{1}{3}\left(\frac{1 - 5^k}{1 - 5}\right) \longrightarrow\) \(\displaystyle \textcolor{blue}{\text{this series diverges since} \ 5 > 1}\).
\(\displaystyle \sum_{k=0}^{\infty}\frac{5^k}{3} = \frac{1}{3} + \frac{5}{3} + \frac{5^2}{3} + \cdots\)
Well this is a geometric series.
\(\displaystyle \sum_{k=0}^{\infty}\frac{5^k}{3} = \lim_{k\rightarrow \infty}\frac{1}{3}\left(\frac{1 - 5^k}{1 - 5}\right) \longrightarrow\) \(\displaystyle \textcolor{blue}{\text{this series diverges since} \ 5 > 1}\).
logistic_guy
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