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Chapter 9: Problem 5

Show that the series \(1 / 1^{2}+1 / 2^{3}+1 / 3^{2}+1 / 4^{3}+\cdots\) is convergent, but that both the Ratio and the Root Tests fail to apply.

Short Answer

Expert verified
The given series is convergent, as determined by the Comparison Test. However, both the Ratio and Root Tests fail to apply because their relevant limit expressions do not exist due to the alternating nature of the series exponent, making the tests inconclusive.

Step by step solution

01

Apply Comparison Test

The Comparison Test can be applied to this series to determine its convergence. This test states that if \( 0\leq a_n \leq b_n \) for all n greater than some positive integer N, and if the series \( \sum_{n=1}^{\infty} b_n \) is convergent, then the series \( \sum_{n=1}^{\infty} a_n \) is also convergent. Here, taking \( b_n = \frac{1}{n^2} \), it can be shown that \( 0\leq a_n\leq b_n \) for all n. The series \( \sum_{n=1}^{\infty} \frac{1}{n^2} \) is a p-series with p=2, and is convergent. Hence, by the Comparison Test, the given series is also convergent.
02

Apply Ratio Test

The Ratio Test is used to determine the convergence or divergence of a series. This test states that if the limit as n approaches infinity of the absolute value of the ratio of consecutive terms \( a_{n+1}/a_n \) of a series is less than 1, the series is convergent, if it is greater than 1, the series is divergent, and if it is equal to 1, the test is inconclusive. Applying this test to the given series, we have \( \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| \). Considering that the series alternates between p=2 and p=3, it is apparent that this limit does not exist. Hence, the Ratio Test is inconclusive and fails to apply.
03

Apply Root Test

The Root Test is used to determine the convergence or divergence of a series. This test states that if the limit as n approaches infinity of the nth root of the absolute value of the nth term \( |a_n|^{1/n} \) of a series is less than 1, the series is convergent, if it is greater than 1, the series is divergent, and if it is equal to 1, the test is inconclusive. Applying this test to the given series, we have \( \lim_{n \to \infty} \left|a_n\right|^{1/n} \). Considering that the series alternates between p=2 and p=3, it is clear that this limit does not exist. Hence, the Root Test is inconclusive and fails to apply.

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Most popular questions from this chapter

If the partial sums \(s_{n}\) of \(\sum_{n=1}^{\infty} a_{n}\) are bounded, show that the series \(\sum_{n=1}^{\infty} a_{n} / n\) converges to \(\sum_{n=1}^{\infty} s_{n} / n(n+1)\)

Discuss the series whose \(n\) th term is: (a) \((-1)^{n} \frac{n^{n}}{(n+1)^{n+1}}\). (b) \(\frac{n^{n}}{(n+1)^{n+1}}\), (c) \((-1)^{n} \frac{(n+1)^{n}}{n^{n}}\) (d) \(\frac{(n+1)^{n}}{n^{n+1}}\).

If \(\left(a_{n}\right)\) is a decreasing sequence of strictly positive aumbers and if \(\sum a_{n}\) is convergent, show that \(\lim \left(n a_{n}\right)=0\)

If \(\sum a_{n}\) is an absolutely convergent series. then the series \(\sum a_{n} \sin n x\) is absolutely and uniformly convergent.

If the partial sums of \(\sum a_{n}\) are bounded, show that the series \(\sum_{n=1}^{\infty} a_{n} e^{-n t}\) converges for \(t>0\).
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