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Chapter 10: Problem 42

Which of the series in Exercises \(17-56\) converge, and which diverge? Use any method, and give reasons for your answers. $$ \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} $$

Short Answer

Expert verified
The series converges by the Ratio Test.

Step by step solution

01

Identify the Series

We are given the series \( \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} \). We need to determine if this series converges or diverges.
02

Consider the Exponential Term

Note that the term \( e^{n} \) grows exponentially as \( n \to \infty \). Exponential growth tends to overwhelm polynomial growth in the numerator and denominator.
03

Apply the Ratio Test

For the series \( \sum a_n = \sum \frac{\ln n}{\sqrt{n} e^{n}} \), apply the ratio test: compute \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
04

Simplify the Ratio of Terms

Compute \( \frac{a_{n+1}}{a_n} = \frac{\ln(n+1)}{\sqrt{n+1} e^{n+1}} \times \frac{\sqrt{n} e^{n}}{\ln n} \). This simplifies to \( \frac{\ln(n+1)}{\ln n} \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \cdot \frac{1}{e} \).
05

Evaluate the Limit

Evaluate the limit: \( L = \lim_{n \to \infty} \frac{\ln(n+1)}{\ln n} \cdot \frac{\sqrt{n}}{\sqrt{n+1}} \cdot \frac{1}{e} \).- The first term \( \frac{\ln(n+1)}{\ln n} \to 1 \) as \( n \to \infty \).- The second term \( \frac{\sqrt{n}}{\sqrt{n+1}} \to 1 \) as \( n \to \infty \).- The third term is \( \frac{1}{e} \).Thus, \( L = \frac{1}{e} < 1 \).
06

Conclusion from Ratio Test

Since the ratio limit \( L < 1 \), by the ratio test, the series \( \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} \) converges.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ratio Test
The ratio test is a powerful tool for determining the convergence or divergence of an infinite series. If you have a series \( \sum a_n \), the test tells you to examine \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \). This is the limit of the ratio of successive terms in the series. Here's what to do once you've computed this limit:
  • If the limit \( L < 1 \), the series converges absolutely.
  • If the limit \( L > 1 \), or if the limit is infinity, the series diverges.
  • If the limit equals 1, the test is inconclusive.
This simplicity makes the ratio test a favorite when dealing with series that include factorials or exponential expressions. In our exercise, we found that \( L = \frac{1}{e} < 1 \). This clearly indicates convergence of the series \( \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} \).
Exponential Growth
Exponential growth refers to increases in quantity by multiplying repeatedly by a constant factor over regular intervals. In mathematical expressions, this is often seen as \( e^n \) or other exponential terms. Such growth is incredibly rapid, especially as \( n \to \infty \).

Exponential growth can overpower other functions like polynomials or logarithms because:
  • It increases much faster than polynomial growth.
  • It plays a significant role in determining the behavior of a series as in long-term terms it dominates over slower-growing functions.
In our series \( \sum_{n=1}^{\infty} \frac{\ln n}{\sqrt{n} e^{n}} \), the exponential term \( e^n \) has an overarching growth effect, rendering the polynomial and logarithmic components relatively insignificant as \( n \to \infty \).
Polynomial Growth
Polynomial growth describes functions where the increase in value is proportional to the power of the variable, like \( n^k \). Examples of polynomial growth include quadratic or cubic functions like \( n^2 \) or \( n^3 \). These functions grow polynomially and are generally slower compared to exponential growth.

In terms of convergence considerations:
  • Polynomial growth becomes important when comparing to other rates of growth, such as exponential growth.
  • Compared to exponential functions, polynomials grow more slowly, making them less influential in series that include terms with exponential growth.
In our series, the logarithmic and square root terms grow polynomially, but their growth is minor compared to the overpowering exponential term \( e^n \). This makes their effect less significant in terms of overall series convergence or divergence.

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Most popular questions from this chapter

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{1}=1, \quad a_{n+1}=a_{n}+\frac{1}{5^{n}} $$

Which of the series in Exercises 13 46 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series' convergence or divergence.) $$ \sum_{n=1}^{\infty} \frac{8 \tan ^{-1} n}{1+n^{2}} $$

Use a CAS to perform the following steps for the sequences. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit \(L\) ? b. If the sequence converges, find an integer \(N\) such that \(\quad\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) $$ a_{n}=\sin n $$

The \(n\) th root of \(n !\) a. Show that \(\lim _{n \rightarrow \infty}(2 n \pi)^{1 /(2 n)}=1\) and hence, using Stirling's approximation (Chapter \(8,\) Additional Exercise 52 \(\mathrm{a}\) , that $$ \sqrt[n]{n !} \approx \frac{n}{e} \quad \text { for large values of } n $$ b. Test the approximation in part (a) for \(n=40,50,60, \ldots\) as far as your calculator will allow.

For a sequence \(\left\\{a_{n}\right\\}\) the terms of even index are denoted by \(a_{2 k}\) and the terms of odd index by \(a_{2 k+1} .\) Prove that if \(a_{2 k} \rightarrow L\) and \(a_{2 k+1} \rightarrow L,\) then \(a_{n} \rightarrow L\)
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