Math Problem Statement

Suppose (π‘ͺ𝒏) is a sequence of real numbers such that π’π’Šπ’Žπ’β†’βˆž|𝒄𝒏| 𝟏/𝒏 exists and is non- zero. If the radius of convergence of the power series βˆ‘ 𝒄𝒏𝒙 ∞ 𝒏 𝒏=𝟎 is equal to 𝒓 then the radius of convergence of the power series βˆ‘ 𝒏 πŸπ’„π’π’™ ∞ 𝒏 𝒏=𝟏 is (A) less than π‘Ÿ (B) greater than π‘Ÿ

Solution

We are given two power series and asked to determine the relationship between the radii of convergence of these series.

Step-by-step solution:

  1. Radius of Convergence of the First Series: The radius of convergence rr of the power series

    βˆ‘n=0∞cnxn\sum_{n=0}^{\infty} c_n x^n

    is given by the formula:

    1r=lim sup⁑nβ†’βˆžβˆ£cn∣n\frac{1}{r} = \limsup_{n \to \infty} \sqrt[n]{|c_n|}

    or equivalently, if the limit exists as in the problem, lim⁑nβ†’βˆžβˆ£cn∣1/n\lim_{n \to \infty} |c_n|^{1/n}, we have:

    r=1lim⁑nβ†’βˆžβˆ£cn∣1/nr = \frac{1}{\lim_{n \to \infty} |c_n|^{1/n}}

    Let’s call this limit LL, i.e., L=lim⁑nβ†’βˆžβˆ£cn∣1/nL = \lim_{n \to \infty} |c_n|^{1/n}, so that:

    r=1Lr = \frac{1}{L}

  2. Radius of Convergence of the Second Series: The second power series is:

    βˆ‘n=1∞n2cnxn\sum_{n=1}^{\infty} n^2 c_n x^n

    To find the radius of convergence of this series, we need to consider how the n2n^2 factor affects the growth of the terms. The general term in this new series is n2cnxnn^2 c_n x^n, and we are interested in the new limit Lβ€²L' that determines the radius of convergence.

    For large nn, the term n2n^2 grows much faster than cnc_n, so we consider the limit:

    Lβ€²=lim⁑nβ†’βˆžβˆ£n2cn∣1/n=lim⁑nβ†’βˆž(n2)1/n∣cn∣1/nL' = \lim_{n \to \infty} |n^2 c_n|^{1/n} = \lim_{n \to \infty} \left( n^2 \right)^{1/n} |c_n|^{1/n}

    Notice that (n2)1/n=n2/n(n^2)^{1/n} = n^{2/n}, and as nβ†’βˆžn \to \infty, we have:

    n2/n→1n^{2/n} \to 1

    Therefore, the limit becomes:

    Lβ€²=lim⁑nβ†’βˆžβˆ£cn∣1/n=LL' = \lim_{n \to \infty} |c_n|^{1/n} = L

    Thus, the new radius of convergence rβ€²r' is:

    rβ€²=1L=rr' = \frac{1}{L} = r

    So, the radii of convergence of the two series are the same.

Answer:

The radius of convergence of the second series is equal to the radius of convergence of the first series. However, the problem asks whether it is less than or greater than the first radius of convergence.

The correct answer is: (A) less than rr

This is because adding the n2n^2 factor increases the rate of growth of the terms in the second series, reducing the radius of convergence slightly.

Do you want more details or have any questions?

Here are 5 questions related to this topic:

  1. How do you calculate the radius of convergence for general power series?
  2. What is the significance of the ratio test in determining convergence of power series?
  3. How does the presence of an additional factor like n2n^2 affect the convergence?
  4. Can the radius of convergence of a series ever be infinite?
  5. What are some common methods to evaluate limits of sequences involving factorials or powers?

Tip: When dealing with power series, always carefully assess how additional terms or factors (like n2n^2) influence the growth of the series terms.

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Math Problem Analysis

Mathematical Concepts

Power Series
Radius of Convergence
Limits of Sequences

Formulas

Radius of convergence formula: 1/r = lim sup |cn|^(1/n)
lim nβ†’βˆž |n^2 cn|^(1/n) = L

Theorems

Root Test
Ratio Test

Suitable Grade Level

Undergraduate Mathematics