If f(x) is continuous, positive and decreasing function on \([k,\infty]\) and \(f(n)=a_n\) then

If \(\int_k^\infty f(x)dx\) is convergent then \(\sum_{n=k}^\infty a_n\) is also convergent.

If \(\int_k^\infty f(x)dx\) is divergent then \(\sum_{n=k}^\infty a_n\) is also divergent.

Given that

\(\sum_{n=1}^\infty3^{-n}\)

Here \(a_n=3^{-n}\)

Since \(a_n=3^{-n}\) so functon f(x) is

\(f(x)=3^{-x}\)

F(x) is positive and decreasing so integral test is applicable for the given series.

Now evaluate \(\int_k^\infty f(x)dx\) so

\(\int_1^\infty3^{-x}dx=[-\frac{3^{-x}}{\ln3}]_1^\infty\)

\(=-\frac{3^{-\infty}}{\ln3}+\frac{3^{-1}}{\ln3}\)

\(=0+\frac{1}{3\ln3}\)

\(=\frac{1}{3\ln3}\)

Since \(\int_1^\infty3^{-x}dx\) is convergent. Hence using integral test \(\sum_{n=1}^\infty3^{-n}\) is also convergent.