dstefan
07-01-2008, 05:22 PM
Chapter 4, Problem 7
Find the radius of convergence R of the power series
\sum_{k=2}^{\infty} \frac{x^k}{k \ln(k)}
and investigate what happens at the points x where |x| = R.
Solution:
It is easy to see that
\label{ch4_ex7_series}
\sum_{k=2}^{\infty} \frac{x^k}{k ~\ln(k)} ~=~ \sum_{k=2}^{\infty} a_k x^k, with a_k = \frac{1}{k \ln(k)}, k > 1.
Note that
\lim_{k \to \infty} |a_k|^{1/k} ~=~ \lim_{k \to \infty} ~\left( \frac{1}{k \ln(k)} \right)^{1/k} ~=~ 1
Recall that, if \lim_{k \to \infty} |a_k|^{1/k} exists, then the radius of convergence of the series \sum_{k=2}^{\infty} a_k x^k is given by R ~=~ \frac{1}{\lim_{k \to \infty} |a_k|^{1/k}}. Then, the radius of convergence of the series (\ref{ch4_ex7_series}) is R = 1, and we conclude that the series is convergent if |x| < 1, and not convergent if |x| > 1.
If x = - 1, the series becomes \sum_{k=1}^{\infty} \frac{(-1)^k}{k \ln(k)}. Since the terms \frac{(-1)^k}{k \ln(k)} have alternating signs and decrease in absolute value to 0, the series is convergent.
If x=1, the series becomes \sum_{k=1}^{\infty} \frac{1}{k \ln(k)}, which was shown to be divergent is Problem 6 from Chapter 4.
Find the radius of convergence R of the power series
\sum_{k=2}^{\infty} \frac{x^k}{k \ln(k)}
and investigate what happens at the points x where |x| = R.
Solution:
It is easy to see that
\label{ch4_ex7_series}
\sum_{k=2}^{\infty} \frac{x^k}{k ~\ln(k)} ~=~ \sum_{k=2}^{\infty} a_k x^k, with a_k = \frac{1}{k \ln(k)}, k > 1.
Note that
\lim_{k \to \infty} |a_k|^{1/k} ~=~ \lim_{k \to \infty} ~\left( \frac{1}{k \ln(k)} \right)^{1/k} ~=~ 1
Recall that, if \lim_{k \to \infty} |a_k|^{1/k} exists, then the radius of convergence of the series \sum_{k=2}^{\infty} a_k x^k is given by R ~=~ \frac{1}{\lim_{k \to \infty} |a_k|^{1/k}}. Then, the radius of convergence of the series (\ref{ch4_ex7_series}) is R = 1, and we conclude that the series is convergent if |x| < 1, and not convergent if |x| > 1.
If x = - 1, the series becomes \sum_{k=1}^{\infty} \frac{(-1)^k}{k \ln(k)}. Since the terms \frac{(-1)^k}{k \ln(k)} have alternating signs and decrease in absolute value to 0, the series is convergent.
If x=1, the series becomes \sum_{k=1}^{\infty} \frac{1}{k \ln(k)}, which was shown to be divergent is Problem 6 from Chapter 4.