Name-Zainab-Bibi-BC200400692-Assignmeng-No-2-Mth-621-solution-a-n-n-n-2-a-n-n-n-2-n-n-2-2-by-ratio-test-a-n-a-n-n-n-2-

Question Number 194851 by Guriya last updated on 17/Jul/23

Name Zainab Bibi

BC 200400692

You can't use 'macro parameter character #' in math mode

Mth 621

solution . .

a_{n} = \frac{(n -)!}{{(n +)}^{2}}

a_{n +} = \frac{(n + -)!}{{(n + +)}^{2}} = \frac{(n)!}{{(n + 2)}^{2}}

by ratio test

\frac{a_{n +}}{a_{n}} = \frac{\frac{(n)!}{(n + 2)}}{\frac{(n -)!}{{(n +)}^{2}}} = \frac{(n)!}{{(n + 2)}^{2}} \times \frac{{(n +)}^{2}}{(n +)!}

li \underset{n \to \infty}{m} \frac{a_{n +}}{a_{n}} = li \underset{n \to \infty}{m} \frac{n (n -)!}{{(n + 2)}^{2}} \times \frac{{(n +)}^{2}}{(n -)!} ∵ (n)! = n (n -)!

li \underset{n \to \infty}{m} \frac{a_{n +}}{a_{n}} = li \underset{n \to \infty}{m} \frac{n {(n +)}^{2}}{{(n + 2)}^{2}} = li \underset{n \to \infty}{m} \frac{n (n^{2} + 1 + 2 n)}{n^{2} + 4 + 4 n}

li \underset{n \to \infty}{m} \frac{(n^{3} + n + {2 n}^{2})}{n^{2} + 4 + 4 n} = li \underset{n \to \infty}{m} \frac{(1 + \frac{1}{n^{2}} + \frac{2}{n})}{\frac{1}{n} + \frac{1}{n^{3}} + \frac{4}{n^{2}}}

Divided by n^{3} to numerator and denominator

Now by Applying limit

\frac{(1 + \frac{1}{\infty^{2}} + \frac{2}{\infty})}{\frac{1}{\infty} + \frac{4}{\infty^{3}} + \frac{4}{\infty^{2}}} = \frac{1 + 0 + 0}{0 + 0 + 0} = \frac{1}{0} = \infty

li \underset{n \to \infty}{m} \frac{a_{n + 1}}{a_{n}} = \infty

Commented by Frix last updated on 17/Jul/23

\frac{1}{0} = \infty is not allowed; instead use this :

\frac{a_{n +}}{a_{n}} = \frac{\frac{(n)!}{(n + 2)}}{\frac{(n -)!}{{(n +)}^{2}}} = \frac{(n)!}{{(n + 2)}^{2}} \times \frac{{(n +)}^{2}}{(n -)!} =

= n \times \frac{n^{2} + 2 n + 1}{n^{2} + 4 n + 4} = n \times (1 + \frac{2 n + 1}{n^{2} + 4 n + 1})

\lim_{n \to \infty} n = \infty

\lim_{n \to \infty} 1 = 1

\lim_{n \to \infty} \frac{2 n + 1}{n^{2} + 4 n + 1} = 0

\infty \times (1 + 0) = \infty

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