Given that:

The series is \(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}\)

By using,

Geometric series test :

A geometric series \(\sum_{n=0}^\infty a_1r^n\text{ or }\sum_{n=1}^\infty a_1r^{n-1}\) is converges if and only if - 1 < r < 1 and its sum is,

\(\sum_{n=1}^\infty a_1r^{n-1}=\frac{a_1}{1-r}\)

Consider the series,

\(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}=7\sum_{n=2}^\infty\frac{(-3)^n}{5^n}\)

\(=7\sum_{n=2}^\infty(\frac{-3}{5})^n\)

The series \(\sum_{n=2}^\infty(\frac{-3}{5})^n\) is the geometric series with first term \((a_1)\) is \(\frac{9}{25}\) and common ratio \((r)=-\frac35\)

Also, \(|r|=|-\frac35|\leq1\)

Then,

By the geometric series,

The series \(\sum_{n=2}^\infty(\frac{-3}{5})^n\) is converges and its sum,

\(\sum_{n=2}^\infty(\frac{-3}{5})^n=\frac{\frac{9}{25}}{1-(-\frac35)}=\frac{9(5)}{25(8)}=\frac{9}{40}\)

Then,

\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}=7\sum_{n=2}^\infty(\frac{-3}{5})^n=7(\frac{9}{40})=\frac{63}{40}\)

Therefore,

The series \(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}\) is converges and its sum is \(\frac{63}{40}\)

The series is \(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}\)

By using,

Geometric series test :

A geometric series \(\sum_{n=0}^\infty a_1r^n\text{ or }\sum_{n=1}^\infty a_1r^{n-1}\) is converges if and only if - 1 < r < 1 and its sum is,

\(\sum_{n=1}^\infty a_1r^{n-1}=\frac{a_1}{1-r}\)

Consider the series,

\(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}=7\sum_{n=2}^\infty\frac{(-3)^n}{5^n}\)

\(=7\sum_{n=2}^\infty(\frac{-3}{5})^n\)

The series \(\sum_{n=2}^\infty(\frac{-3}{5})^n\) is the geometric series with first term \((a_1)\) is \(\frac{9}{25}\) and common ratio \((r)=-\frac35\)

Also, \(|r|=|-\frac35|\leq1\)

Then,

By the geometric series,

The series \(\sum_{n=2}^\infty(\frac{-3}{5})^n\) is converges and its sum,

\(\sum_{n=2}^\infty(\frac{-3}{5})^n=\frac{\frac{9}{25}}{1-(-\frac35)}=\frac{9(5)}{25(8)}=\frac{9}{40}\)

Then,

\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}=7\sum_{n=2}^\infty(\frac{-3}{5})^n=7(\frac{9}{40})=\frac{63}{40}\)

Therefore,

The series \(\sum_{n=2}^\infty\frac{7\cdot(-3)^n}{5^n}\) is converges and its sum is \(\frac{63}{40}\)