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Chapter 9: Problem 35

Use the Comparison Test or Limit Comparison Test to determine whether the following series converge. $$\sum_{k=1}^{\infty} \frac{1}{2 k-\sqrt{k}}$$

Short Answer

Expert verified
Answer: The series $\sum_{k=1}^{\infty} \frac{1}{2k-\sqrt{k}}$ diverges by the Limit Comparison Test.

Step by step solution

01

Identify the Series to Compare With

We will compare the given series with the harmonic series: $$\sum_{k=1}^{\infty} \frac{1}{k}$$ The harmonic series is a known divergent series.
02

Apply the Limit Comparison Test

We will use the Limit Comparison Test, which requires calculating the limit of the ratio of the terms in the two series as n approaches infinity. We calculate the limit: $$\lim_{k \to \infty} \frac{\frac{1}{2k-\sqrt{k}}}{\frac{1}{k}}$$ This limit simplifies to: $$\lim_{k \to \infty} \frac{k}{2k-\sqrt{k}}$$
03

Simplify the Limit Expression

To simplify the limit expression, we can divide both the numerator and the denominator by k: $$\lim_{k \to \infty} \frac{1}{2-\frac{\sqrt{k}}{k}}$$
04

Evaluate the Limit

Now we can find the limit as k approaches infinity: $$\lim_{k \to \infty} \frac{1}{2-\frac{\sqrt{k}}{k}} = \frac{1}{2-0} = \frac{1}{2}$$ Since the limit is a positive finite number, by the Limit Comparison Test, both series (the given series and the harmonic series) will have the same convergence behavior.
05

Determine Convergence of the Series

Since we compared our given series with the harmonic series, which is a divergent series, and their limit ratio is a positive finite number, it means that the given series also diverges. So, the series $$\sum_{k=1}^{\infty} \frac{1}{2k-\sqrt{k}}$$ diverges by the Limit Comparison Test.

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Most popular questions from this chapter

Determine whether the following series converge or diverge. $$\sum_{k=1}^{\infty} \frac{2^{k}+3^{k}}{4^{k}}$$

Consider the sequence \(\left\\{F_{n}\right\\}\) defined by $$F_{n}=\sum_{k=1}^{\infty} \frac{1}{k(k+n)^{\prime}}$$ for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ). for \(n=0,1,2, \ldots . .\) When \(n=0,\) the series is a \(p\) -series, and we have \(F_{0}=\pi^{2} / 6\) (Exercises 65 and 66 ).

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