Problem evaluating the limit to find the radius of convergence

[brunette15]

Hi everyone,
I am trying to evaluate the radius of convergence for the following power series: (k!(x-1)^k)/((2^k)(k^k))

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

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

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

 

[Prove It]

Hi everyone,
I am trying to evaluate the radius of convergence for the following power series: (k!(x-1)^k)/((2^k)(k^k))

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

So far i have L = lim k--> inf (k^k(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 $\displaystyle \begin{align*} 2\mathrm{e} \end{align*}$.

 

[brunette15]

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!