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The comparasion Test:

\( \sum \frac{1}{2^{n}} \) Convergent \( \sum \frac{1}{2^{n}+1} \)

\( \sum \frac{1}{2^{n}+1}<\sum \frac{1}{2^{n}} \) Basic Comparision test     \( \sum \text { an } \& \sum b n \)

                                                                    \( \sum a n  \ \& \sum bn \)

 P- Series        \( \sum \frac{1}{n^{p}} \)            \( p>1 \)        Convergent

                                                   \( p \leq 1 \)        divergent

② Geometric series    \( \sum a r^{n-1} \)         \( |r|<1 \)         convergent

                                                              \( |r| \geq | \)        Divergent

Determine whether the series \( \sum_{n=1}^{\infty} \frac{5}{2 n^{2}+4 n+3} \) converges or diverges.

\( \ln (x) \ll x^{n} \ll a^{x} \leq n ! \ll x^{x} \)                    \( 2 n^{2}+4 n+3 \)

\( \frac{5}{2 n^{2}+4 n+3}<\frac{5}{2 n^{2}} \)

\( a n<b n \)

\( \sum \frac{5}{2 n^{2}}=\frac{5}{2} \sum \frac{1}{n^{2}} \)

P- series with \( p=2>1 \rightarrow \) Convergent

\( \sum \frac{5}{2 n^{2}+4 n+3} \) is Convergent by Basic comperasion Test

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

as \( n \rightarrow \infty \quad a=\frac{1}{n} \)

\( \frac{1}{n-\ln (n)}>\frac{1}{n} \) 

\( \sum \frac{1}{n} \) is divergent -P series \(P=1\)

\( \sum \frac{1}{n-\ln (n)} \) is divergent by Basic Comparasion test.

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