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Chapter 9: Problem 10

Give an example of a divergent series \(\sum a_{n}\) with \(\left(a_{n}\right)\) decreasing and such that \(\lim \left(n a_{n}\right)=0\).

Short Answer

Expert verified
An example of a divergent series that satisfies the given conditions is \(\sum_{n=1}^{\infty} \frac{1}{n 2^n}\).

Step by step solution

01

Understand the definition of a divergent series

A divergent series is a series that does not converge; that is, a series for which the sequence of partial sums does not have a finite limit. A famous example of a divergent series is the harmonic series. However, this does not fully satisfy our conditions as the limit of \(n*a_{n}\) for the harmonic series does not equal 0.
02

Find an appropriate series

One of the examples that meets the given conditions is the series \(\sum_{n=1}^{\infty} \frac{1}{n}\). Notwithstanding, it's well-known that this series diverges as it's a harmonic series. However, to fully satisfy our condition, we need to modify this series slightly.
03

Modify the series

We modify the harmonic series slightly by including a decreasing exponential factor. One possible choice is \(\sum_{n=1}^{\infty} \frac{1}{n 2^n}\). This series is a decreasing sequence since each \(a_{n} = \frac{1}{n 2^n}\) is less than the previous one and it's a divergent series.
04

Verify the conditions

Now, we have to confirm whether the sequence \(n*a_{n}\) approaches 0 as \(n\) approaches to infinity. For our sequence, \(n*a_{n} = \frac{n}{n 2^n} = \frac{1}{2^n}\). As \(n\) tends to infinity, \(\frac{1}{2^n}\) approaches 0. Hence, \(\lim_{n\rightarrow \infty} n*a_{n} = 0\), satisfying our conditions.

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Most popular questions from this chapter

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 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 \(\sum_{n=1}^{\infty} s_{n} / n(n+1)\)

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