show that Σ_(n=2) ^∞ ((n−4)/(n^2 −2n+1)) is converge or diverge by ussing any test


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 136559 by mohammad17 last updated on 23/Mar/21

$s h o w t h a t \underset{n = 2}{\sum^{\infty}} \frac{n - 4}{n^{2} - 2 n + 1} i s c o n v e r g e o r d i v e r g e$ $b y u s s i n g a n y t e s t$
Commented by mr W last updated on 23/Mar/21

$i t h i n k s,$ $i f f (n) i s a p o l y n o m i a l o f d e g r e e p$ $a n d g (n) i s a p o l y n o m i a l o f d e g r e e q,$ $t h e n \underset{n = 1}{\sum^{\infty}} \frac{f (n)}{g (n)} c o n v e r g e s i f q - p ⩾ 2,$ $o t h e r w i s e i t d i v e r g e s .$
	Answered by physicstutes last updated on 23/Mar/21

	$Alternatively$ $a_{n} = \frac{1}{n} and b_{n} = \frac{n - 4}{n^{2} - 2 n + 1}$ $\frac{b_{n}}{a_{n}} = \frac{n^{2} - 4 n}{n^{2} - 2 n + 1}; \lim_{n \to \infty} (\frac{n^{2} - 4 n}{n^{2} - 2 n + 1}) = 1 \Rightarrow Σ b_{n} and Σ a_{n} either they both$ $converge or they both diverge, but we established that the p - series$ $\underset{n = 2}{\sum^{\infty}} \frac{1}{n} = Σ a_{n} diverges since its p value ⩽ 1, hence Σ b_{n} or \underset{n = 2}{\sum^{\infty}} \frac{n - 4}{n^{2} - 2 n + 1}$ $diverges$
	Answered by Olaf last updated on 23/Mar/21

	$\frac{n - 4}{n^{2} - 2 n + 1} \underset{\infty}{\sim} \frac{1}{n}$ $But the harmonic serie diverges$ $\Rightarrow \underset{n = 2}{\sum^{\infty}} \frac{n - 4}{n^{2} - 2 n + 1} diverges$
	Answered by physicstutes last updated on 23/Mar/21

	$when n \to \infty we can get a comparison series of \underset{n = 2}{\sum^{\infty}} \frac{1}{n} which is divergnt$ $let b_{n} = \frac{n - 4}{n^{2} - 2 n + 1} and a_{n} = \frac{1}{n}$ $since Σ a_{n} diverges, \Rightarrow Σ b_{n} diverges by the comparison test$
	Answered by physicstutes last updated on 23/Mar/21

	$another alternative$ $\underset{n = 2}{\sum^{\infty}} (\frac{n - 4}{n^{2} - 2 n + 1}) = \underset{n = 2}{\sum^{\infty}} [\frac{n - 4}{{(n - 1)}^{2}}] = \underset{n = 2}{\sum^{\infty}} (\frac{1}{n - 1} - \frac{3}{{(n - 1)}^{2}})$ $\underset{n = 2}{\sum^{\infty}} (\frac{1}{n - 1}) another divergent harmonic$ $\underset{n = 2}{\sum^{\infty}} [\frac{3}{{(n - 1)}^{2}}] yet another divergent one so \underset{n = 2}{\sum^{\infty}} (\frac{n - 4}{n^{2} - 2 n + 1}) is divergent$ $from all universes .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum