Determining Convergence or Divergence
Which of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.
∑ (from n=1 to ∞) (1 − n) / n2ⁿ

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First, write down the general term of the series: $a_n = \frac{1 - n}{n 2^n}$.

Observe the behavior of the terms as $n$ becomes large. Since \$2^n$ grows exponentially and $n$ grows linearly, the denominator grows very fast compared to the numerator.

To determine convergence, consider applying the Ratio Test, which is useful for series involving terms with exponential components. The Ratio Test states to compute $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$.

Calculate $\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(1 - (n+1)) / ((n+1) 2^{n+1})}{(1 - n) / (n 2^n)} \right|$ and simplify this expression step-by-step.

Evaluate the limit $L$ as $n$ approaches infinity. If $L < 1$, the series converges absolutely; if $L > 1$, it diverges; if $L = 1$, the test is inconclusive and other methods should be considered.

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주요 개념

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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.

Ratio Test

The Ratio Test is used to determine the convergence of a series by examining the limit of the absolute value of the ratio of consecutive terms. If this limit is less than 1, the series converges absolutely; if greater than 1, it diverges; if equal to 1, the test is inconclusive.

Behavior of Exponential Terms in Series

Exponential terms like 2^n in the denominator grow very rapidly, often causing terms to approach zero quickly. This rapid decay can lead to convergence, especially when combined with polynomial terms in the numerator or denominator.

Determining Convergence or DivergenceWhich of the series in Exercises 17–56 converge, and which diverge? Use any method, and give reasons for your answers.∑ (from n=1 to ∞) (1 − n) / n2ⁿ