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Chapter 9: Problem 30

Determine the convergence or divergence of the following series. $$\sum_{k=2}^{\infty} \frac{k^{e}}{k^{\pi}}$$

Short Answer

Expert verified
Answer: The series diverges.

Step by step solution

01

Rewrite the series

We can rewrite the given series as $$\sum_{k=2}^{\infty} \frac{k^{e}}{k^{\pi}}= \sum_{k=2}^{\infty} k^{e-\pi}$$
02

Choose a simple series to compare

We will compare our given series with the simple series $$\sum_{k=2}^{\infty} k^{-1}$$ which is a harmonic series and diverges.
03

Apply the limit comparison test

Let's find the limit of the ratio of our given series term with the simple series term as k approaches infinity: $$\lim_{k\to\infty}\frac{k^{e-\pi}}{k^{-1}} = \lim_{k\to\infty}k^{e-\pi+1}$$
04

Find the exponent e - 蟺 + 1

In our limit, we have the exponent e - 蟺 + 1. Let's find out if it's positive, negative, or zero: $$e - \pi + 1 = 2.718 - 3.142 + 1 \approx 0.576$$ Since the exponent is positive, we proceed with the limit.
05

Find the limit

Since the exponent is positive, the limit becomes $$\lim_{k\to\infty}k^{0.576} = \infty$$
06

Determine convergence or divergence

Since the result of the limit comparison test is infinite (not finite and positive), we cannot conclude that our series behaves similarly to the harmonic series. However, since both series have positive terms and the terms of our given series become arbitrarily large as k approaches infinity, we can conclude that the given series diverges. If the terms of our series were decreasing and k would approach infinity, we could argue for convergence. But in this case, the series does not meet these conditions. Therefore, the given series $$\sum_{k=2}^{\infty} \frac{k^{e}}{k^{\pi}}$$ diverges.

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Most popular questions from this chapter

A glimpse ahead to power series Use the Ratio Test to determine the values of \(x \geq 0\) for which each series converges. $$\sum_{k=1}^{\infty} \frac{x^{k}}{2^{k}}$$

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