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Chapter 9: Problem 13

(a) Does the series \(\sum_{n=1}^{\infty}\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}\right)\) converge? (b) Does the series \(\sum_{n=1}^{\infty}\left(\frac{\sqrt{n+1}-\sqrt{n}}{n}\right)\) converge?

Short Answer

Expert verified
The series \(\sum_{n=1}^{\infty}\left(\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}}\right)\) diverges and the series \(\sum_{n=1}^{\infty}\left(\frac{\sqrt{n+1}-\sqrt{n}}{n}\right)\) converges.

Step by step solution

01

Simplify the series expressions

First, simplify the series expressions. It should be noted that \(\sqrt{n+1}-\sqrt{n}\) can be simplified according to its conjugate, which leads to a simpler equation. For the first expression, it could be rewritten as \[ \sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n+1}+\sqrt{n}}\right)\] and the second expression as \[ \sum_{n=1}^{\infty}\left(\frac{1}{n(\sqrt{n+1}+\sqrt{n})}\right)\]
02

Identifying Convergence for the first series

Now, consider the first series. A key point is to note that \(\frac{1}{\sqrt{n+1}+\sqrt{n}} > \frac{1}{2\sqrt{n+1}}\) since adding \(\sqrt{n}\) to the denominator on the left hand side makes it bigger - hence the fraction, smaller. So, one can compare the series to \( \sum_{n=1}^{\infty}\frac{1}{2\sqrt{n+1}} \), which is a divergent series. By the comparison test, the first series is divergent.
03

Identifying Convergence for the second series

For the second series, here it is seen as: \[ \sum_{n=1}^{\infty}\left(\frac{1}{n(\sqrt{n+1}+\sqrt{n})}\right) \]. Simplifying, \[ \sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n+1}+\sqrt{n}}\right) \cdot \frac{1}{n} \]. The key identity here is that \(\frac{1}{n} > \frac{1}{n+1}\). This allows one to write an equivalent series as \[ \sum_{n=1}^{\infty}\left(\frac{1}{2\sqrt{n+1}}\right) \cdot \frac{1}{n} \], which can then be compared to a convergent series like \[ \sum_{n=1}^{\infty}\left(\frac{1}{n^{3/2}}\right) \]. Hence, by the comparison test, this series converges.

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Most popular questions from this chapter

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

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 \(t>0\).

Show that the radius of convergence \(R\) of the power series \(\sum a_{n} x^{n}\) is given by \(\lim \left(\left|a_{n} / a_{n+1}\right|\right)\) whenever this limic exists. Give an example of a power series where this limit does not exist.

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.
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