Determining Convergence or Divergence
Which of the series in Exercises 13–46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.)
∑ (from n=1 to ∞) eⁿ / (1 + e²ⁿ)

Guida verificata passo dopo passo

First, write down the general term of the series: \(a_n = \frac{e^n}{1 + e^{2n}}\).

Simplify the term \(a_n\) by factoring the denominator: notice that \(e^{2n} = (e^n)^2\), so rewrite \(a_n\) as \(\frac{e^n}{1 + (e^n)^2}\).

Divide numerator and denominator by \(e^{2n}\) to express \(a_n\) in a form that reveals its behavior as \(n \to \infty\): \(a_n = \frac{e^n / e^{2n}}{(1 / e^{2n}) + 1} = \frac{e^{-n}}{1 + e^{-2n}}\).

Analyze the limit of \(a_n\) as \(n\) approaches infinity: since \(e^{-n} \to 0\) and \(e^{-2n} \to 0\), the term \(a_n\) approaches 0.

To determine convergence, compare \(a_n\) to a known convergent series: since \(a_n\) behaves like \(e^{-n}\) for large \(n\), and the series \(\sum e^{-n}\) is a geometric series with ratio less than 1, use the Comparison Test or Limit Comparison Test to conclude about convergence.

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Infinite Series and Convergence

An infinite series is the sum of infinitely many terms. Determining whether a series converges means checking if the sum approaches a finite limit as the number of terms grows indefinitely. If the sum does not approach a finite value, the series diverges.

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The Comparison Test helps determine convergence by comparing the given series to another series with known behavior. If the terms of the given series are smaller than those of a convergent series, it also converges; if larger than a divergent series, it diverges.

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