What is the radius of convergence of #sum_1^oo ((2x)^n ) / 8^(2n)#?

Answer 1

32

#sum_1^oo ((2x)^n ) / 8^(2n)# = #sum_1^oo ((2x) / 8^(2))^n# = #sum_1^oo ((x) / 32)^n#

Using the radius definition, we can end here, but just to be thorough:

#sum_1^oo ((x) / 32)^n = lim_(n->oo) (1-(x/32)^n)/(1-x/32)#
Which converges only if #|x/32| < 1 => |x| < 32#

Consequently, 32 is the convergence ratio.

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Answer 2

The radius of convergence ( R ) of the series ( \sum_{n=1}^{\infty} \frac{(2x)^n}{8^{2n}} ) can be found using the ratio test:

[ R = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

Where ( a_n = \frac{(2x)^n}{8^{2n}} ).

Applying the ratio test, we get:

[ \lim_{n \to \infty} \left| \frac{\frac{(2x)^{n+1}}{8^{2(n+1)}}}{\frac{(2x)^n}{8^{2n}}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(2x)^{n+1} \cdot 8^{2n}}{(2x)^n \cdot 8^{2(n+1)}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{(2x)^{n+1} \cdot 8^{2n}}{(2x)^n \cdot 8^{2n+2}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{2x}{64} \right| ]

[ = \frac{|2x|}{64} ]

For the series to converge, this ratio must be less than 1:

[ \frac{|2x|}{64} < 1 ]

[ |2x| < 64 ]

[ |x| < 32 ]

Therefore, the radius of convergence is ( R = 32 ).

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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.

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.

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.

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.

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