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• Notes

The comparasion Tests 2:

$\sum \frac{1}{n+\ln(n)}<\sum \frac{1}{n}$

$(p-\text { series with } p=1 \ divergent)$             Basic comparasion Test

Test Fails

$\sum \text { an } \& \sum bn$                                                 Limit Comparasion Test

$\lim _{n \rightarrow \infty} \frac{a n}{b n}=c \quad(0, \infty)$

The comparasion Tests 2:

Limit Comparasion test:-LCT

$\sum_{n=1}^{\infty} a_{n} \ \& \sum_{n=1}^{\infty} b n$ are Positive term Series

if $\lim _{n \rightarrow \infty} \frac{a n}{b n}=c \quad \in \quad(0, \infty)$

$∴\sum \text { an } \& \sum bn$ are both Convergent

or divergent.

if $\lim _{n \rightarrow \infty} \frac{a n}{b n}=0, \infty \longrightarrow$Bastc comparasion Test.

Determine whether the series $\sum \frac{1}{n+\ln (n)}$ is convergent or divergent.

Positive term Series $\longrightarrow$  Limit Comparasion Test (LCT)

$\lim _{n \rightarrow \infty} \frac{a n}{b n}=\lim \frac{\frac{1}{n+\ln (n)}}{\frac{1}{n}}=\lim _{n \rightarrow \infty} \frac{n}{n+\ln (n)}=\frac{\infty}{\infty}$

Apply L'Hopital Rule:

$\lim _{n \rightarrow \infty}\left(\frac{a n}{b n}\right)=\lim _{n \rightarrow \infty} \frac{1}{1+\frac{1}{n}}=\frac{1}{1+\frac{1}{\infty}}=1=c>0 \in(0, \infty)$

since $\sum \mathrm{bn}$ is Divergent

$∴ \sum a_{n} \text { is Divergent using }(L C T)$

Determine whether the series $\sum_{n=1}^{\infty} \sin \left(\frac{1}{n}\right)$ is convergent or divergent.

positive term Series $\longrightarrow$ Limit comparasion Test (L C T)

$bn=\frac{1}{n}$
$\lim _{n \rightarrow \infty} \frac{a n}{b n}=\lim _{n \rightarrow \infty} \frac{\sin \left(\frac{1}{n}\right)}{\frac{1}{n}}=\frac{\sin \left(\frac{1}{\infty}\right)}{\frac{1}{\infty}}=\frac{0}{0}$                                        $\frac{ \sin (0)}{0}=\frac{0}{0}$

Apply L'Hopital Rule:

$\lim _{n \rightarrow \infty} \frac{a n}{b n}=\lim _{n \rightarrow \infty} \frac{-\frac{1}{n^{2}} \cdot \cos \frac{1}{n}}{-\frac{1}{n^{2}}}=\lim _{n \rightarrow \infty} \cos \frac{1}{n}=\cos \left(\frac{1}{\infty}\right)=\cos (0)=1$                $\in(0, \infty)$

but $\sum_{n=1}^{\infty} \frac{1}{n}$ is divergent (P- series with P=1)

$∴ \sum_{n=1}^{\infty} \sin ( \frac{1}{n})$ is divergent too.