Find the sum of the first nterms of the G.P 3+1+(1/3)+...and show that the sum cannot exceed (9/2) however great n may be.


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Question Number 46697 by KMA last updated on 30/Oct/18

$F i n d t h e s u m o f t h e f i r s t n t e r m s$ $o f t h e G . P 3 + 1 + \frac{1}{3} + . . . a n d s h o w t h a t$ $t h e s u m c a n n o t e x c e e d \frac{9}{2} h o w e v e r$ $g r e a t n m a y b e .$
	Answered by Joel578 last updated on 30/Oct/18

	$S_{n} = \frac{a (1 - r^{n})}{1 - r} = \frac{3 (1 - {(\frac{1}{3})}^{n})}{1 - \frac{1}{3}}$ $= \frac{9}{2} (1 - {(\frac{1}{3})}^{n})$ $\lim_{n \to \infty} S_{n} = \lim_{n \to \infty} [\frac{9}{2} (1 - {(\frac{1}{3})}^{n})]$ $= \frac{9}{2} . \lim_{n \to \infty} (1 - {(\frac{1}{3})}^{n})$ $= \frac{9}{2} (1 - 0)$ $= \frac{9}{2}$ $It is showed that as n goes larger and larger,$ $its value getting closer to \frac{9}{2} and never exceed \frac{9}{2}$
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