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

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Identify the general term of the series: \(a_k = \frac{1}{\sqrt{k} \times e^{\sqrt{k}}}\).

Consider the behavior of the term \(a_k\) as \(k \to \infty\). Since \(e^{\sqrt{k}}\) grows very rapidly, the terms \(a_k\) approach zero.

To determine convergence, apply the Comparison Test by comparing \(a_k\) to a simpler series. Note that \(e^{\sqrt{k}}\) grows faster than any polynomial, so compare \(a_k\) to \(\frac{1}{e^{\sqrt{k}}}\).

Since \(\sum \frac{1}{e^{\sqrt{k}}}\) converges (because the terms decrease exponentially), and \(a_k < \frac{1}{e^{\sqrt{k}}}\) for all large \(k\), by the Comparison Test, the original series converges.

Alternatively, you can apply the Integral Test by considering the integral of the function \(f(x) = \frac{1}{\sqrt{x} e^{\sqrt{x}}}\) from 1 to infinity and showing that it converges, which implies the series converges.

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주요 개념

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Convergence of Infinite Series

An infinite series converges if the sequence of its partial sums approaches a finite limit. Determining convergence involves analyzing the behavior of the terms as the index grows large, ensuring the sum does not diverge to infinity.

Comparison and Limit Comparison Tests

These tests compare the given series to a known benchmark series. If terms of the given series behave similarly to a convergent or divergent series, we can conclude the same about the original series, simplifying convergence analysis.

Exponential and Root Functions in Series Terms

Understanding how exponential functions like e^(√k) grow faster than polynomial or root functions is crucial. This growth rate often dominates the denominator, causing terms to decrease rapidly, which influences the convergence of the series.

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