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Ch. 10 - Infinite Sequences and Series
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 10, Problema 10.5.46

Determining Convergence or Divergence
In Exercises 17–46, use any method to determine whether the series converges or diverges. Give reasons for your answer.
∑(from n=3 to ∞) [2n² / n²ⁿ]

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First, write down the general term of the series: \(a_n = \frac{2n^2}{n^{2n}}\) for \(n \geq 3\).
Observe the form of \(a_n\) to decide which convergence test is appropriate. Since the term involves \(n\) raised to a power that depends on \(n\), consider using the Root Test or the Ratio Test.
Apply the Root Test by computing \(\lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\frac{2n^2}{n^{2n}}}\).
Simplify the expression inside the limit: \(\sqrt[n]{2n^2} = (2n^2)^{1/n}\) and \(\sqrt[n]{n^{2n}} = n^{2n/n} = n^2\). So the limit becomes \(\lim_{n \to \infty} \frac{(2n^2)^{1/n}}{n^2}\).
Evaluate the limit: as \(n \to \infty\), \((2n^2)^{1/n} \to 1\) because the \(n\)th root of any polynomial grows slowly, while \(n^2\) in the denominator grows without bound. Therefore, the limit is \(0\), which is less than \(1\), indicating that the series converges by the Root Test.

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