Home > Calculus > Determining the Radius and Interval of Convergence for a Power Series

What is the interval of convergence of #sum_{k=0}^oo 2^(k) / ((2k)!* x^(k)) #?

Answer 1

Aurora Alvarado

#sum_{k=0}^oo 2^(k) / ((2k)!* x^(k))# converges #forall x in RR, x ne 0#

We are aware of that

#e^x=sum_{k=0}^oox^k/(k!)# and also #(e^x+e^-x)/2 = cosh(x) = sum_{k=0}^oox^{2k}/(2k!)#

but

#sum_{k=0}^oo 2^(k) / ((2k)!* z^(k))=sum_{k=0}^oo(2/z)^k/(2k!)=sum_{k=0}^oo(sqrt(2/z))^{2k}/(2k!)#

so for #x > 0#

#sum_{k=0}^oo(sqrt(2/x))^{2k}/(2k!)=cosh(sqrt(2/x))#

Analogous considerations can be stated for #x < 0#

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Answer 2

Charles Arocha

The interval of convergence for the series ∑ (2^k / ((2k)! * x^k)) is determined by applying the ratio test. By using the ratio test, you can evaluate the limit as k approaches infinity of the absolute value of the ratio of consecutive terms. This limit must be less than 1 for convergence. After simplifying and evaluating this limit, you will find that the interval of convergence is -∞ < x < ∞.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.