Ask question

# 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.

### Relevant Questions asked 2021-01-19
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.
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{3}^{{-{n}}}$$ asked 2021-02-11
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.
$$\frac{\ln2}{2}+\frac{\ln3}{3}+\frac{\ln4}{4}+\frac{\ln5}{5}+\frac{\ln6}{6}+...$$ asked 2021-01-16

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.
$$\sum_{n=1}^\infty\frac{1}{(2n+3)^3}$$ asked 2021-01-06

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.
$$\sum_{n=1}^\infty\frac{1}{n+3}$$ asked 2021-01-16

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.
$$\sum_{n=2}^\infty\frac{1}{n\sqrt{\ln n}}$$ asked 2020-10-18

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.
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{\left({2}{n}+{3}\right)}^{{3}}}}}$$ asked 2021-03-08

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.
$$\displaystyle{\sum_{{{n}={1}}}^{\infty}}{\frac{{{1}}}{{{n}+{3}}}}$$ asked 2021-02-08

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.
$$\displaystyle{\sum_{{{n}={2}}}^{\infty}}{\frac{{{1}}}{{{n}\sqrt{{{\ln{{n}}}}}}}}$$ asked 2020-11-08

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.
$$\sum_{n=1}^\infty\frac{\arctan n}{n^2+1}$$ asked 2021-02-09

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.
$$\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...$$

...