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.

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.