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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

Prove by Induction that the function given by \(f(x):=e^{-1 / x^{2}}\) for \(x \neq 0, f(0):=0\), has derivatives of all orders at every point and that all of these derivatives vanish at \(x=0\). Hence this function is not given by its Taylor expansion about \(x=0\).

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 \(a\) and \(b\) are positive numbers, then \(\sum(a n+b)^{-p}\) converges if \(p>1\) and diverges if \(p \leq 1\).

For \(n \in \mathbb{N}\), let \(c_{n}\) be defined by \(c_{n}:=\frac{1}{1}+\frac{1}{2}+\cdots+1 / n-\ln n .\) Show that \(\left(c_{n}\right)\) is a decreasing sequence of positive numbers. The limit \(C\) of this sequence is called Euler's Constant and is approximately equal to \(0.577 .\) Show that if we put $$b_{n}:=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\cdots-\frac{1}{2 n}$$ then the sequence \(\left(b_{n}\right)\) converges to \(\ln 2 .\) [Hint: \(\left.b_{n}=c_{2 n}-c_{n}+\ln 2 .\right\\}\)

The preceding exercise may fail if the terms are not positive. For example, let \(a_{1 j}:=+1\) if \(i-j=1, a_{i j}:=-1\) if \(i-j=-1\), and \(a_{i j}:=0\) elsewhere. Show that the iterated sums $$\sum_{i=1}^{\infty} \sum_{j=1}^{\infty} a_{i j} \quad \text { and } \quad \sum_{j=1}^{\infty} \sum_{i=1}^{\infty} a_{i j}$$ both exist but are not equal.
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