Test the following series for convergence: (i) ∑x^n/n (ii) ∑x^n/n² (iii) ∑(sin 1/n) x^n (iv) ∑anx^n, an = { 2^−n n = 2k 3^−n n = 2k + 1.

Question

Test the following series for convergence:

[latex](i) \sum{\frac{x^{n}}{n}} (ii) \sum{\frac{x^{n}}{n^{2}}} (iii) \sum{\left(\sin \frac{1}{n} ight)x^{n}}[/latex]

[latex](iv) \sum{a_{n}x^{n}}, a_{n} = \begin{cases} 2^{−n} & n = 2k \ 3^{−n} & n = 2k + 1. \end{cases}[/latex]

Accepted Answer

(i) With [latex]a_{n} = 1/n,[/latex] we have

[latex]R = \lim \left|\frac{a_{n}}{a_{n+1}}\right| = \lim \frac{n + 1}{n} = 1.[/latex]

At x = 1, we obtain the harmonic series [latex]\sum{1/n},[/latex] which we know is divergent. x = −1 yields the alternating series [latex]\sum(−1)^{n}/n,[/latex] which is convergent. Thus the series converges on [−1, 1) and diverges on [latex]\mathbb{R}\setminus [−1, 1).[/latex]

(ii) In this case [latex]a_{n} = 1/n^{2}[/latex] and

[latex]R = \lim \frac{(n + 1)^{2}}{n^{2}} = 1.[/latex]

The series converges at x = ±1, hence it converges on the closed interval [−1, 1] and diverges outside it.

(iii)

[latex]R = \lim \frac{\sin(1/n)}{\sin(1/(n + 1))} = 1.[/latex]

When x = −1, the alternating series [latex]\sum{(−1)^{n}} \sin(1/n)[/latex] 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

[latex]\sin \frac{1}{n} ≥ \frac{2}{nπ}[/latex] for all [latex]n ∈ \mathbb{N}.[/latex]

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

[latex]\underset{k→∞}{\lim} a_{2k}^{1/2k} = \frac{1}{2}, \underset{k→∞}{\lim} a_{2k+1}^{1/(2k+1)} = \frac{1}{3},[/latex]

we have [latex]\lim \sup a_{n}^{1/n} = 1/2,[/latex] hence R = 2. At x = ±2, the n-th term [latex]a_{n}x^{n}[/latex] 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.

Question 9.13: Test the following series for convergence: (i) ∑x^n/n (ii) ∑...

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