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Chapter 9: Problem 34

Determine convergence or divergence for each of the series. Indicate the test you use. $$ \sum_{n=1}^{\infty} \frac{n}{2+n 5^{n}} $$

Short Answer

Expert verified
The series converges by the ratio test.

Step by step solution

01

Simplify the General Term

First, let's simplify the general term for the series: \( \frac{n}{2+n 5^n} \). The denominator grows very quickly because it includes \( 5^n \), which is an exponential term. For large values of \( n \), \( n 5^n \) dominates over the constant 2.
02

Apply the Ratio Test (asymptotically)

Consider using the ratio test to determine the convergence of the series. The ratio test involves finding the limit:\[L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|,\]where \( a_n = \frac{n}{2+n 5^n} \). Calculate the ratio:\[\frac{a_{n+1}}{a_n} = \frac{\frac{n+1}{2+(n+1)5^{n+1}}}{\frac{n}{2+n5^n}}.\]Simplifying this involves a complex setup but noting the dominance of \( 5^n \), notice that asymptotically:\[ \frac{n5^n}{(n+1)5^{n+1}} \approx \frac{1}{5}\]This implies \( L = \frac{1}{5} < 1 \), suggesting the series converges by the ratio test.
03

Conclusion

Based on the result from the ratio test, we have \( L < 1 \), which tells us that the series \( \sum_{n=1}^{\infty} \frac{n}{2+n 5^n} \) 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 popular method to determine the convergence or divergence of an infinite series. If you have a series in the form of \( \sum a_n \), the Ratio Test can tell us whether it converges absolutely, diverges, or the test is inconclusive.
  • Start by considering the limit \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \).
  • If \( L < 1 \), the series converges absolutely.
  • If \( L > 1 \) or \( L = \infty \), the series diverges.
  • If \( L = 1 \), the Ratio Test is inconclusive, and you鈥檒l need another test to make a conclusion.
In the exercise, by using the Ratio Test, the calculated limit for the series \( \sum \frac{n}{2+n 5^n} \) was found to be \( \frac{1}{5} \), which is less than 1. Hence, the series converges. This result effectively tells us that as \( n \) becomes very large, the terms of the series become smaller at a faster than linear pace, leading to convergence.
Exponential Growth
Exponential growth is a powerful mathematical concept that describes processes increasing over time, wherein the growth rate is proportional to the value of the function.
For example, in this series, the term \( 5^n \) contributes to exponential growth by rapidly increasing as \( n \) increases. Here are some characteristics:
  • Exponential terms usually overpower linear or polynomial terms, especially as \( n \) becomes very large.
  • In the context of the series \( \sum \frac{n}{2+n 5^n} \), the rapid growth of \( 5^n \) makes its influence dominant in the denominator, ensuring that the fraction as a whole shrinks toward zero as n increases.
Understanding exponential growth is crucial for recognizing why certain series converge or diverge. When such growth is present, it鈥檚 often the case that exponential growth in the denominator will result in the sequence of terms of the series diminishing quickly to zero, thus supporting convergence.
Series Simplification
Simplifying a series is a necessary step when you first approach problems related to series convergence or divergence. It often involves identifying and focusing on the dominant terms that heavily influence the behavior of the series.
In the series \( \frac{n}{2 + n5^n} \), the term \( 5^n \) is the most significant contributor in the denominator, especially for large \( n \). Here are some considerations for simplification:
  • Ignore lower order terms when they don鈥檛 significantly impact the behavior of the series at large \( n \). In this case, "2" becomes negligible compared to \( n5^n \).
  • Focus on terms that dictate the growth or decline speed. For example, since \( 5^n \) grows extremely fast, it confirms the decrease in terms like \( \frac{n}{5^n} \) to near zero, promoting convergence.
By simplifying, you can better apply tests like the Ratio Test, and gain insights into the series鈥� behavior, making it easier to deduce the ultimate convergence or divergence outcome.

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Most popular questions from this chapter

Let \(y=y(x)=x-\frac{\lambda}{3 !}+\frac{\lambda}{5 !}-\frac{\lambda}{7 !}+\cdots .\) Show that \(y\) satisfies the differential equation \(y^{\prime \prime}+y=0\) with the conditions \(y(0)=0\) and \(y^{\prime}(0)=1\). From this, guess a simple formula for \(y\).

Tom and Joel are good runners, both able to run at a constant speed of 10 miles per hour. Their amazing dog Trot can do even better; he runs at 20 miles per hour. Starting from towns 60 miles apart, Tom and Joel run toward each other while Trot runs back and forth between them. How far does Trot run by the time the boys meet? Assume that Trot started with Tom running toward Joel and that he is able to make instant turnarounds. Solve the problem two ways. (a) Use a geometric series. (b) Find a shorter way to do the problem.

$$ \text { Prove: If } \sum_{k=1}^{\infty} a_{k} \text { diverges, so does } \sum_{k=1}^{\infty} c a_{k} \text { for } c \neq 0 \text { . } $$

Show that the positive terms of the alternating harmonic series form a divergent series. Show the same for the negative terms.

One can sometimes find a Maclaurin series by the method of equating coefficients. For example, let $$ \tan x=\frac{\sin x}{\cos x}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots $$ Then multiply by \(\cos x\) and replace \(\sin x\) and \(\cos x\) by their series to obtain $$ \begin{aligned} x-\frac{x^{3}}{6}+\cdots &=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots\right)\left(1-\frac{x^{2}}{2}+\cdots\right) \\ &=a_{0}+a_{1} x+\left(a_{2}-\frac{a_{0}}{2}\right) x^{2}+\left(a_{3}-\frac{a_{1}}{2}\right) x^{3}+\cdots \end{aligned} $$ Thus, $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}-\frac{a_{0}}{2}=0, \quad a_{3}-\frac{a_{1}}{2}=-\frac{1}{6}, \quad \cdots $$ so $$ a_{0}=0, \quad a_{1}=1, \quad a_{2}=0, \quad a_{3}=\frac{1}{3}, \quad \cdots $$ and therefore $$ \tan x=0+x+0+\frac{1}{3} x^{3}+\cdots $$ which agrees with Problem 1 . Use this method to find the terms through \(x^{4}\) in the series for \(\sec x\).
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