• عربي

Need Help?

Subscribe to Calculus B

###### \${selected_topic_name}
• Notes

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.