How to find the sum of this series ? $$\sum_{n=1}^∞ \Big(\frac{4}{9}\Big)^n $$
In General,
Does the series $\sum_{n=1}^∞ k^n$ , where $k$ is a random real number, have the finite sum ?
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Chinnapparaj R
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MAJ
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Do you mean $n$ in your sums? Your lower indices on the sum are $i$, so any terms in $n$ just come out of the sum.... – PhysicsMathsLove Dec 04 '17 at 10:28
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ow, sorry, a typo – MAJ Dec 04 '17 at 10:29
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There is no $i$ in the summed term $(4/9)^n$,which isn't how sum notation is usually defined. – coffeemath Dec 04 '17 at 10:29
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I think you should have $i$ somwhere in the body of the summation – Vinyl_cape_jawa Dec 04 '17 at 10:29
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sorry, the i is meant to be n, already edited. – MAJ Dec 04 '17 at 10:32
1 Answers
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If you are talking about $\sum_{n=1}^{\infty} (\frac{4}{9})^n$ then you see its a geometric series with common ratio of $\frac{4}{9}$ which is less than 1 guaranteeing convergence of the series.
Thus for $\sum_{n=0}^{\infty} a r^n = \frac{a}{1-r}$.
So
$\sum_{n=1}^{\infty} (\frac{4}{9})^n = \frac{\frac{4}{9}}{1 - \frac{4}{9}} = \frac{4}{5}$.
BAYMAX
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2Well the same remark as to previous answer: there is no need to answer, when it is just the formula for the sum of geometric progression. A link to wikipedia is enough. – Nikita Evseev Dec 04 '17 at 10:37
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