 # 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}+... naivlingr 2021-02-09 Answered

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.
$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+...$

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Given: The series
$\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+...$
To determine: The convergence and divergence of the given series by Integral test.
Explanation:
Integral Test: Let f(x) be a positive, continuous and decreasing function for $x\in \left[1,\mathrm{\infty }\right)$ and $f\left(n\right)={a}_{n}$ where n is any natural number then the improper integral ${\int }_{1}^{\mathrm{\infty }}f\left(x\right)dx$ and infinite $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}$ both converge or diverge simultaneously.
Here the infinite sum is $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+...=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n+1}$
Therefore ${a}_{n}=\frac{1}{2n+1}$
So let $f\left(x\right)=\frac{1}{2x+1}$ for $x\in \left[1,\mathrm{\infty }\right)$
Then clearly f(x) is continuous, positive, and decreasing for $x\in \left[1,\mathrm{\infty }\right)$ and $f\left(n\right)=\frac{1}{2n+1}={a}_{n}$ for any natural number n.
Hence Integral Test can be applied for $f\left(x\right)=\frac{1}{2x+1}$ and ${a}_{n}\frac{1}{2n+1}$
Now checking for the convergence or divergence of the improper integral
${\int }_{1}^{\mathrm{\infty }}f\left(x\right)dx$ for $f\left(x\right)=\frac{1}{2x+1}$
So,
${\int }_{1}^{\mathrm{\infty }}f\left(x\right)dx$
$=\underset{t\to \mathrm{\infty }}{lim}{\int }_{1}^{t}\frac{1}{2x+1}dx$
$=\frac{1}{2}\underset{t\to \mathrm{\infty }}{lim}{\int }_{1}^{t}\frac{2}{2x+1}dx$ [multiplying and dividing by 2]
$=\frac{1}{2}\underset{t\to \mathrm{\infty }}{lim}\left[\mathrm{ln}\left(2x+1\right){\right]}_{1}^{t}\phantom{\rule{1em}{0ex}}\left[\because \int \frac{{f}^{\prime }\left(x\right)}{f\left(x\right)}dx=\mathrm{ln}\left[f\left(x\right)\right]\right]$
$=\frac{1}{2}\underset{t\to \mathrm{\infty }}{lim}\left[\mathrm{ln}\left(2t+1\right)-\mathrm{ln}3\right]$
$=\mathrm{\infty }$
Hence ${\int }_{1}^{\mathrm{\infty }}f\left(x\right)dx={\int }_{1}^{\mathrm{\infty }}\frac{1}{2x+1}dx$ diverges
Therefore by Integral test
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n+1}=\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+...$ also diverge.
Answer: The series $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}+...$ is divergent

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