what will be the value of x^1/2.x^1/4.x^1/8... to infinity
\[x^\frac{ 1 }{ 2 },x ^\frac{ 1}{ 4 },x^\frac{ 1 }{ 8 }.......\]
a.x^2 b.x c.x^3/2 d.x^3 e.none of these
whats the exact question ?
are you adding the terms ?
here the series could be in AP,GP or HP .....sum of infinite series is the question i think
wolfram says the sum does not converge http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5E%7B%5Cinfty%7D+%28x%5E%281%2F2%5Ek%29%29
book shows ans as\[x\]
do you have solution ?
no, no hint no solution, therefore i post here
consider x = 1
is 1 + 1 + 1 + ... = 1 ?
lol
question is wrong, if possible take a snapshot and attach
ok wait
for example, when x = 3 the sum of first 3 terms itself give u answer very much greater than x http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bk%3D1%7D%5E%7B3%7D+%283%5E%281%2F2%5Ek%29%29
its not a series, its a product man !
\[\large \sum_{n = 1}^{\infty}x^{2^{-n}}\]
He he he.. :)
\[\huge x^{ \frac{1}{2} +\frac{1}{4} +\frac{1}{8} +\cdots }\]
This looks nice..
oh ,the author was ambigous
It is a GP with first term 1/2 and common ratio 1/2
author was clear, you made it ambiguous by replacing \(\huge \cdot \) by a \(\large ,\)
ohh !!!
work out the sum in exponent
That dot (.) represents Product.. :)
the previous question all had (,)..... so i thought ,lol
stating the problem as clearly as possible and writing your thoughts on the problem really helps
can you work the sum in exponent?
but thanks for that comma ,i have to be cautious ....lol
:)
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