Problem evaluating the limit to find the radius of convergence

In summary, the participants are discussing how to evaluate the radius of convergence for a given power series. They mention using the ratio test on the absolute value series to determine where the series converges absolutely, and solving for x to find the radius of convergence. The final result is a radius of convergence of 2e.
  • #1
brunette15
58
0
Hi everyone,
I am trying to evaluate the radius of convergence for the following power series: (k!(x-1)k)/((2k)(kk))

I have begun by trying to compute L = lim k-->inf (an+1/an). To then be able to say R = 1/L.

So far i have L = lim k--> inf (kk(k+1)!)/(2(k+1)k+1k!)

From here i am having trouble evaluating the limit however :/
 
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  • #2
brunette15 said:
Hi everyone,
I am trying to evaluate the radius of convergence for the following power series: (k!(x-1)k)/((2k)(kk))

I have begun by trying to compute L = lim k-->inf (an+1/an). To then be able to say R = 1/L.

So far i have L = lim k--> inf (kk(k+1)!)/(2(k+1)k+1k!)

From here i am having trouble evaluating the limit however :/

You actually have to use the ratio test on the ABSOLUTE VALUE series, to show where the series converges ABSOLUTELY, as you know nothing about what happens for particular x values, and the ratio test only works for positive term series.

Anyway, you need to evaluate the limit of the ratio of terms of the absolute value series, and then find which x values allow this limit to be less than 1. This will guarantee convergence. In other words, we have to solve this for x...

$\displaystyle \begin{align*} \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{\frac{(k+1)!(x-1)^{k+1}}{2^{k+1}(k+1)^{k+1}}}{\frac{k!(x-1)^k}{2^kk^k}} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{(k+1)! (x-1)^{k+1} 2^k k^k}{k!(x-1)^k 2^{k + 1}(k + 1)^{k + 1}} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{(k+1) (x - 1) k^k}{2(k + 1)^{k + 1}} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{(x - 1) k^k}{2(k + 1)^k} \right| &< 1 \\ \frac{ \left| x - 1 \right| }{2} \lim_{k \to \infty} \left( \frac{k}{k + 1} \right) ^k &< 1 \\ \left| x - 1 \right| \lim_{ k \to \infty} \mathrm{e}^{\ln{ \left[ \left( \frac{k}{k+1} \right) ^k \right] }} &< 2 \\ \left| x - 1 \right| \mathrm{e}^{\lim_{k \to \infty} k \ln{ \left( \frac{k}{k + 1} \right) }} &< 2 \\ \left| x - 1 \right| \mathrm{e}^{ \lim_{k \to \infty} \frac{\ln{(k)} - \ln{(k + 1)}}{\frac{1}{k}} } &< 2 \\ \left| x - 1 \right| \mathrm{e}^{ \lim_{k \to \infty} \frac{\frac{1}{k} - \frac{1}{k + 1}}{-\frac{1}{k^2}} } &< 2 \textrm{ by L'Hospital's Rule} \\ \left| x - 1 \right| \mathrm{e}^{\lim_{k \to \infty} -k^2 \left( \frac{k + 1 - k}{k(k + 1)} \right) } &< 2 \\ \left| x - 1 \right| \mathrm{e}^{ \lim_{k \to \infty} -\frac{k}{k + 1} } &< 2 \\ \left| x - 1 \right| \mathrm{e}^{ \lim_{k \to \infty} \frac{1}{k} - 1 } &< 2 \\ \left| x - 1 \right| \mathrm{e}^{-1} &< 2 \\ \left| x - 1 \right| &< 2\mathrm{e} \end{align*}$

So the radius of convergence is $\displaystyle \begin{align*} 2\mathrm{e} \end{align*}$.
 
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  • #3
Prove It said:
You actually have to use the ratio test on the ABSOLUTE VALUE series, to show where the series converges ABSOLUTELY, as you know nothing about what happens for particular x values, and the ratio test only works for positive term series.

Anyway, you need to evaluate the limit of the ratio of terms of the absolute value series, and then find which x values allow this limit to be less than 1. This will guarantee convergence. In other words, we have to solve this for x...

$\displaystyle \begin{align*} \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{\frac{(k+1)!(x-1)^{k+1}}{2^{k+1}(k+1)^{k+1}}}{\frac{k!(x-1)^k}{2^kk^k}} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{(k+1)! (x-1)^{k+1} 2^k k^k}{k!(x-1)^k 2^{k + 1}(k + 1)^{k + 1}} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{(k+1) (x - 1) k^k}{2(k + 1)^{k + 1}} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{(x - 1) k^k}{2(k + 1)^k} \right| &< 1 \\ \frac{ \left| x - 1 \right| }{2} \lim_{k \to \infty} \left( \frac{k}{k + 1} \right) ^k &< 1 \\ \left| x - 1 \right| \lim_{ k \to \infty} \mathrm{e}^{\ln{ \left[ \left( \frac{k}{k+1} \right) ^k \right] }} &< 2 \\ \left| x - 1 \right| \mathrm{e}^{\lim_{k \to \infty} k \ln{ \left( \frac{k}{k + 1} \right) }} &< 2 \\ \left| x - 1 \right| \mathrm{e}^{ \lim_{k \to \infty} \frac{\ln{(k)} - \ln{(k + 1)}}{\frac{1}{k}} } &< 2 \\ \left| x - 1 \right| \mathrm{e}^{ \lim_{k \to \infty} \frac{\frac{1}{k} - \frac{1}{k + 1}}{-\frac{1}{k^2}} } &< 2 \textrm{ by L'Hospital's Rule} \\ \left| x - 1 \right| \mathrm{e}^{\lim_{k \to \infty} -k^2 \left( \frac{k + 1 - k}{k(k + 1)} \right) } &< 2 \\ \left| x - 1 \right| \mathrm{e}^{ \lim_{k \to \infty} -\frac{k}{k + 1} } &< 2 \\ \left| x - 1 \right| \mathrm{e}^{ \lim_{k \to \infty} \frac{1}{k} - 1 } &< 2 \\ \left| x - 1 \right| \mathrm{e}^{-1} &< 2 \\ \left| x - 1 \right| &< 2\mathrm{e} \end{align*}$

So the radius of convergence is $\displaystyle \begin{align*} 2\mathrm{e} \end{align*}$.

I see! Thankyou!
 

FAQ: Problem evaluating the limit to find the radius of convergence

1. What is the definition of a limit?

A limit is a fundamental concept in calculus that describes the behavior of a function as its input approaches a certain value. It represents the value that a function is approaching, or tending towards, as the input gets closer and closer to a specific point or value.

2. How is a limit used to find the radius of convergence?

When evaluating the limit to find the radius of convergence, we are essentially determining the largest value of x for which the series will converge. This can be done by taking the limit as n approaches infinity of the absolute value of the ratio between the n+1 term and the nth term in the series. If this limit is less than 1, then the series will converge, and the value of x that satisfies this condition is the radius of convergence.

3. What is the significance of the radius of convergence in a power series?

The radius of convergence is an important concept in power series as it determines the interval of values for which the series will converge. If the value of x is within the radius of convergence, then the power series will converge and accurately approximate the function at that point. If the value of x is outside the radius of convergence, then the series will not converge and cannot be used to approximate the function at that point.

4. How can we determine the radius of convergence for a given power series?

The radius of convergence can be determined by analyzing the ratio of the nth term to the (n-1)th term in the series and taking the limit as n approaches infinity. If this limit is less than 1, then the series will converge and the value of x that satisfies this condition is the radius of convergence. This can also be done by using the ratio test or the root test, which are commonly used methods in calculus to determine the convergence of a series.

5. Are there any other methods for finding the radius of convergence?

Yes, there are other methods for finding the radius of convergence, such as the Cauchy-Hadamard theorem and the Taylor series method. These methods involve more advanced mathematical concepts and may be used in specific situations where the ratio or root test may not be applicable. However, for most power series, the ratio or root test is the most commonly used method for determining the radius of convergence.

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