babeeb0oL

2020-10-28

Which of the series, and which diverge?
Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.)
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n-1}$

au4gsf

The given series is
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{2n-1}$
Use Limit Comparison Test to find whether the given series is convergent or divergent.
The nth term of series is
${a}_{n}=\frac{1}{2n-1}$
Consider a series
$\sum _{n=1}^{\mathrm{\infty }}{B}_{n}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}$
then, ${B}_{n}=\frac{1}{n}$
Now,
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{{a}_{n}}{{B}_{n}}\right)=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{\frac{1}{2n-1}}{\frac{1}{n}}\right)$
$\underset{n\to \mathrm{\infty }}{lim}\left(\frac{n}{2n-1}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{n}{n\left(2-\frac{1}{n}\right)}\right)$
$=\underset{n\to \mathrm{\infty }}{lim}\left(\frac{1}{2-\frac{1}{n}}\right)$
$=\frac{1}{2}>0$
Since, series $\left\{{B}_{n}\right\}$ is divergent.
Therefore, the given series is also divergent.

Jeffrey Jordon