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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.

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