Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. sum_{n=1}^infty3^{-n}

Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series. sum_{n=1}^infty3^{-n}

Question
Series
asked 2021-01-05
Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.
\(\sum_{n=1}^\infty3^{-n}\)

Answers (1)

2021-01-06
Integral test:
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.
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