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if the radius of convergence of sum of anXⁿ is R prove that the radius of convergence of sum of anX²n is R¹/²

User GoTop
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Final answer:

The radius of convergence for the modified series ∑anX²ⁿ is the square root of the original radius R, due to the effect of the squared term on the root test used to determine the radius.

Step-by-step explanation:

If the radius of convergence of a power series ∑anX¹ is R, to prove that the radius of convergence of the series ∑anX²ⁿ is √R, we must look at how the radius of convergence is determined for a power series.

The radius of convergence is found using the ratio test or the root test, where, for instance, R = 1/ℓ(ℕ[an]) when the limit exists. In the series ∑anX²ⁿ, each term is “an times X to the power of 2n”.

Applying the ratio or root test to the modified series with terms anX²ⁿ, we are effectively taking the nth root of X to the power of 2n, which simplifies to (X²)ⁿ.

This indicates that the new limit for R will now be based on (√X)ⁿ, which means the new radius of convergence is the square root of the original radius of convergence (i.e., √R). The argument's power within the series terms determines how the radius of convergence is affected.

Therefore, changing the exponent from n to 2n in the series terms scales down the effective change in the radius by a square root. To sum up, if the original radius of convergence is R for the series ∑anXⁿ, the radius of convergence for the series ∑anX²ⁿ is indeed √R.

User Mitya
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