Given the series: 9+frac{117}{4}+frac{1521}{16}+frac{19773}{64}+... does this series converge or diverge? If the series converges, find the sum of the series.

Question
Series
asked 2021-02-03
Given the series:
\(9+\frac{117}{4}+\frac{1521}{16}+\frac{19773}{64}+...\)
does this series converge or diverge? If the series converges, find the sum of the series.

Answers (1)

2021-02-04
To determine
Given:
An infinite series \(9+\frac{117}{4}+\frac{1521}{16}+\frac{19773}{64}+...\)
To determine:
Given series is convergent or divergent.
Calculation
Consider the given series:
\(9+\frac{117}{4}+\frac{1521}{16}+\frac{19773}{64}+...\)
Here, \(a_1=9\)
\(a_2=\frac{117}{4}\)
\(a_3=\frac{1521}{16}\) and so on.
Now, \(\frac{a_2}{a_1}=\frac{\frac{117}{4}}{9}=\frac{13}{4}\)
Similarly, \(\frac{a_3}{a_2}=\frac{\frac{1521}{16}}{\frac{117}{4}}=\frac{13}{4}\)
Similarly, \(\frac{a_4}{a_3}=\frac{\frac{19773}{64}}{\frac{1521}{16}}=\frac{13}{4}\)
Here, \(\frac{a_2}{a_1}=\frac{a_3}{a_2}=\frac{a_4}{a_3}\)
Therefore, given series is geometric series with common ratio \(r=\frac{13}{4}\)
Here, \(r=\frac{13}{4}>1\)
We know, a geometric series with common ratio >1 is always divergent.
Therefore, the given series is divergent.
0

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