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\[\sum_{0}^{\infty} 1\div \left( \sqrt{2} \right)^{n}\]
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did u check if it converges?
no how do i go about that
It's a geometric series, since sqrt(2) > 1, it converges, to a formula you really should look up and memorize.
the one where it says -1<r<1?
and since it is convergent how do i find its sum?
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No, I'll give it: \[\sum_{n=0}^{\ \infty}(\frac{1}{k})^n=\frac{1}{1-\frac{1}{k}}\]
thats it?
Yes.
i thought it was more complicated
Nope. Pretty straight-forward.
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wow thank you
where is the good answer button haha?
Right side of the tan bar with your name in it.
thanks
Any time.
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