Find-the-sum-to-n-terms-of-the-series-S-x-r-0-n-e-x-r-

Question Number 2207 by Yozzi last updated on 08/Nov/15

F i n d t h e s u m t o n t e r m s o f t h e s e r i e s

S (x) = \underset{r = 0}{\sum^{n}} e^{- x^{r}} .

Commented by 123456 last updated on 09/Nov/15

S (x) = \underset{r = 0}{\sum^{n}} e^{- x^{r}} = \frac{1}{e} + \underset{r = 1}{\sum^{n}} e^{- x^{r}} (n > 0)

= \frac{1 + \underset{r = 1}{\sum^{n}} e^{1 - x^{r}}}{e} (n > 0)

= \frac{\underset{r = 0}{\sum^{n}} e^{1 - x^{r}}}{e}

Answered by Rasheed Soomro last updated on 16/Nov/15

S (x) = \underset{r = 0}{\sum^{n}} e^{- x^{r}}

e^{x} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots + \frac{x^{n}}{n!}

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e^{- x^{0}} = e^{- 1} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \dots \pm \frac{1}{n!} [+ f o r e v e n n, - f o r o d d n]

+ e^{- x^{1}} = e^{- x} = 1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots \pm \frac{x^{n}}{n!}

+ e^{- x^{2}} = = 1 - x^{2} + \frac{x^{4}}{2!} - \frac{x^{6}}{3!} + \dots \pm \frac{x^{2 n}}{n!}

+ e^{- x^{3}} = = 1 - x^{3} + \frac{x^{6}}{2!} - \frac{x^{9}}{3!} + \dots . \pm \frac{x^{3 n}}{n!}

+ e^{- x^{4}} = = 1 - x^{4} + \frac{x^{8}}{2!} - \frac{x^{12}}{3!} + \dots \pm \frac{x^{4 n}}{n!}

+ \dots . \dots . \dots . . \dots . \dots . . \dots \dots .

+ \dots . \dots . \dots \dots \dots . \dots . . \dots \dots .

+ e^{- x^{n}}

S (x) i s s u m o f n + 1 g e o m e t r i c s e r i e s

S (x) = (1 + 1 + \dots n + 1 t e r m s) - (1 + x + x^{2} + \dots n + 1 t e r m s)

+ \frac{1}{2!} (1 + x^{2} + x^{4} + \dots n + 1 t e r m s) - \frac{1}{3!} (1 + x^{3} + x^{6} + \dots n + 1 t e r m s)

\dots + \frac{1}{n!} (1 \pm x^{n} + x^{2 n} \dots . n + 1 t e r m s) [+ f o r e v e n n, - f o r o d d]

= 1 (n + 1) - \frac{1 (1 - x^{n + 1})}{1 - x} + \frac{1}{2!} (\frac{1 (1 - {(x^{2})}^{n + 1}}{1 - x^{2}}) \dots .

= (n + 1) - \frac{(1 - x^{n + 1})}{1 - x} + \frac{1}{2!} (\frac{1 - {(x^{2})}^{n + 1}}{1 - x^{2}}) - \dots

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