# 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
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.
$$\sum_{n=1}^\infty3^{-n}$$

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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