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How do you prove that in Calculus that the Sum of (n-5)/(n8^n) from n=5 to infinity is convergent
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is it \[\sum\frac{n-5}{n^n}\]?
\[\sum_{n=5}^{\infty} \frac{ n-5 }{ n8^{n}}\]
Split it into two sums: \[ \frac{n}{n\cdot8^n}-\frac{5}{n\cdot8^n}=\frac{1}{1\cdot8^n}-\frac{5}{n\cdot8^n} \]Note that: \[ \frac{5}{n\cdot8^n}<\frac{5}{8^n} \]And \[ \frac{5}{8^n} \]Converges, hence: \[ \frac{5}{n\cdot8^n} \]Also converges. Since this is the difference of convergent series, the result must converge.
\(\dfrac{n-5}{n}\left(\dfrac{1}{8}\right)^{n} < \dfrac{n}{n}\left(\dfrac{1}{8}\right)^{n} = \left(\dfrac{1}{8}\right)^{n}\) The one on the right is a Geometric Series and is known to converge.
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