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.)

$\sum _{n=1}^{\mathrm{\infty}}\frac{1}{2n-1}$

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.)

au4gsf

Skilled2020-10-29Added 95 answers

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,

$\underset{n\to \mathrm{\infty}}{lim}(\frac{{a}_{n}}{{B}_{n}})=\underset{n\to \mathrm{\infty}}{lim}(\frac{\frac{1}{2n-1}}{\frac{1}{n}})$

$\underset{n\to \mathrm{\infty}}{lim}(\frac{n}{2n-1})$

$=\underset{n\to \mathrm{\infty}}{lim}(\frac{n}{n(2-\frac{1}{n})})$

$=\underset{n\to \mathrm{\infty}}{lim}(\frac{1}{2-\frac{1}{n}})$

$=\frac{1}{2}>0$

Since, series$\left\{{B}_{n}\right\}$ is divergent.

Therefore, the given series is also divergent.

Use Limit Comparison Test to find whether the given series is convergent or divergent.

The nth term of series is

Consider a series

then,

Now,

Since, series

Therefore, the given series is also divergent.

Jeffrey Jordon

Expert2021-12-16Added 2605 answers

Answer is given below (on video)

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