Question 9.13: Test the following series for convergence: (i) ∑x^n/n (ii) ∑...
Test the following series for convergence:
(i) \sum{\frac{x^{n}}{n}} (ii) \sum{\frac{x^{n}}{n^{2}}} (iii) \sum{\left(\sin \frac{1}{n}\right)x^{n}}
(iv) \sum{a_{n}x^{n}}, a_{n} = \begin{cases} 2^{−n} & n = 2k \\ 3^{−n} & n = 2k + 1. \end{cases}
(i) With a_{n} = 1/n, we have
R = \lim \left|\frac{a_{n}}{a_{n+1}}\right| = \lim \frac{n + 1}{n} = 1.
At x = 1, we obtain the harmonic series \sum{1/n}, which we know is divergent. x = −1 yields the alternating series \sum(−1)^{n}/n, which is convergent. Thus the series converges on [−1, 1) and diverges on \mathbb{R}\setminus [−1, 1).
(ii) In this case a_{n} = 1/n^{2} and
R = \lim \frac{(n + 1)^{2}}{n^{2}} = 1.
The series converges at x = ±1, hence it converges on the closed interval [−1, 1] and diverges outside it.
(iii)
R = \lim \frac{\sin(1/n)}{\sin(1/(n + 1))} = 1.
When x = −1, the alternating series \sum{(−1)^{n}} \sin(1/n) results. Since the sequence sin(1/n) decreases monotonically to 0, the series converges at this point. Using the inequality sin x ≥ 2x/π for all x ∈ [0, π/2], which was proved in example 7.11, we obtain
\sin \frac{1}{n} ≥ \frac{2}{nπ} for all n ∈ \mathbb{N}.
Since the series ∑ 2/nπ diverges, so does ∑ sin(1/n) by the comparison test. The interval of convergence is therefore [−1, 1).
(iv) Since
\underset{k→∞}{\lim} a_{2k}^{1/2k} = \frac{1}{2}, \underset{k→∞}{\lim} a_{2k+1}^{1/(2k+1)} = \frac{1}{3},
we have \lim \sup a_{n}^{1/n} = 1/2, hence R = 2. At x = ±2, the n-th term a_{n}x^{n} does not converge to 0, so this series converges on (−2, 2) only.
Uniform convergence for power series is characterized by the following simple result.