Question

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.∑ (from k = 1 to ∞) tan⁻¹(1 / √k)

Accepted Answer

Identify the series given: $$ \sum_{k=1}^{\infty} 	an^{-1}\left( \frac{1}{\sqrt{k}} ight) . We $$want to determine if this infinite series converges or diverges. Recall that for large $$ k $$, $$ 	an^{-1}(x) $$ behaves approximately like $$ x $$ when $$ x  is $$close to zero. Since $$ \frac{1}{\sqrt{k}} 	o 0  as$$ $$ k 	o \infty $$, we can compare $$ 	an^{-1}\left( \frac{1}{\sqrt{k}} ight)  to$$ $$ \frac{1}{\sqrt{k}} . $$Use the Comparison Test or Limit Comparison Test by comparing the given series to the series $$ \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}} $$, which is a p-series with $$ p = \frac{1}{2} . $$Recall that a p-series $$ \sum \frac{1}{k^p} $$ converges if and only if $$ p \u003e 1 . $$Since $$ p = \frac{1}{2} \u003c 1 $$, the series $$ \sum \frac{1}{\sqrt{k}} $$ diverges. Therefore, if the terms of our original series behave like $$ \frac{1}{\sqrt{k}} $$ for large $$ k $$, the original series will also diverge by comparison. Conclude that the series $$ \sum_{k=1}^{\infty} 	an^{-1}\left( \frac{1}{\sqrt{k}} ight) $$ diverges because its terms do not decrease fast enough to produce a convergent sum.

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.
∑ (from k = 1 to ∞)tan⁻¹(1 / √k)

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Key Concepts

Convergence of Infinite Series

Comparison Test and Limit Comparison Test

Behavior of arctan(x) for Small Arguments

11–86. Applying convergence tests Determine whether the following series converge. Justify your answers.∑ (from k = 1 to ∞)tan⁻¹(1 / √k)