/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 34 \(\sum_{n=1}^{\infty} \frac{n}{2... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\sum_{n=1}^{\infty} \frac{n}{2+n 5^{n}}\)

Short Answer

Expert verified
The series converges.

Step by step solution

01

Understanding the Series

The given series is \(\sum_{n=1}^{\infty} \frac{n}{2+n \cdot 5^{n}}\). We need to determine whether this series converges or diverges.
02

Considering the Dominant Term

Notice that for each term, as \(n\) increases, \(n \cdot 5^n\) grows very rapidly because it is exponential. Thus, \(2 + n \cdot 5^n \approx n \cdot 5^n\) for large \(n\).
03

Simplifying the Term

Approximate each term as \(\frac{n}{n \cdot 5^n} = \frac{1}{5^n}\). So, the series can be compared to another series \(\sum_{n=1}^{\infty} \frac{1}{5^n}\), which is a geometric series.
04

Geometric Series Test

A geometric series \(\sum_{n=1}^{\infty} ar^n\) converges if and only if \(|r| < 1\). Here, \(r = \frac{1}{5}\), thus \(|r| < 1\), indicating convergence.
05

Comparison Test

Since \(\frac{n}{2+n \cdot 5^n} \approx \frac{1}{5^n}\) for large \(n\) and \(\frac{1}{5^n}\) is a convergent geometric series, we compare using the Limit Comparison Test. Compute \(\lim_{n \to \infty} \frac{\frac{n}{2 + n \cdot 5^n}}{\frac{1}{5^n}} = \lim_{n \to \infty} \frac{n \cdot 5^n}{2 + n \cdot 5^n} = 1\).
06

Conclusion on Convergence

Since the limit is a positive finite number and the comparison series converges, by the Limit Comparison Test, the original series \(\sum_{n=1}^{\infty} \frac{n}{2 + n \cdot 5^n}\) also converges.

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Key Concepts

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

Limit Comparison Test
The Limit Comparison Test is an essential tool in calculus for determining the convergence or divergence of series. It helps compare an unfamiliar series with a familiar one that we already know the behavior of.

The main idea is simple: if you have two series, say \( \sum a_n \) and \( \sum b_n \), and you suspect they behave similarly, you test them by examining the limit of the ratio of their terms. Specifically, you compute:

\[ \lim_{n \to \infty} \frac{a_n}{b_n}. \]

Then, if this limit is a positive finite number, both series either converge or diverge together.

  • If \( \sum b_n \) converges and the limit is positive and finite, then \( \sum a_n \) converges.
  • If \( \sum b_n \) diverges and the limit is positive and finite, then \( \sum a_n \) also diverges.
In the original problem, we used this test to compare \( \frac{n}{2 + n \cdot 5^n} \) with \( \frac{1}{5^n} \). The positive finite limit confirmed that since \( \sum \frac{1}{5^n} \) converges, our series does too.
Geometric Series
A geometric series is a foundational concept in mathematics, particularly in calculus and series convergence. It is of the form \( \sum_{n=0}^{\infty} ar^n \), where \( a \) is the first term and \( r \) is the common ratio.

  • The convergence of a geometric series depends solely on the value of \( r \).
  • If \( |r| < 1 \), then the series converges to \( \frac{a}{1-r} \).
  • If \( |r| \geq 1 \), then the series diverges.
In the problem at hand, the series \( \sum_{n=1}^{\infty} \frac{1}{5^n} \) was identified as a geometric series with \( a = \frac{1}{5} \) and \( r = \frac{1}{5} \). Since \( |r| < 1 \), this series converges.

By recognizing geometric series, we could apply the concept to benchmark the original series, thus ascertaining its convergence through comparison.
Calculus
Calculus, a branch of mathematics that includes techniques such as differentiation and integration, is fundamental in analyzing series and their convergence. In particular, series and sequences are used to understand functions and their behavior over intervals.

To assess convergence in series, calculus leverages several tests, like the Limit Comparison Test, Ratio Test, and Root Test, among others. Each plays a crucial role in examining infinite sums or extended sequences.

Calculus also introduces key concepts like limits and approximations. In our problem, these concepts helped approximate \( \frac{n}{2 + n \cdot 5^n} \) as \( \frac{1}{5^n} \) for large \( n \), which simplified our comparison using calculus-based tests.

  • Series and sequences utilize limits to analyze behavior as \( n \) approaches infinity.
  • Approximations simplify complex expressions, making problems more manageable.
The principles of calculus ensure that we have a robust framework to tackle series problems and derive conclusions about their behavior in the infinite realm.

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Most popular questions from this chapter

In Problems \(1-8\), find the convergence set for the given power series. $$ \sum_{n=1}^{\infty} \frac{x^{n}}{3^{n}} $$

Use a CAS to find the first four nonzero terms in the Maclaurin series for each of the following. Check Problems 43-48 to see that you get the same answers using the methods of Section 9.7. $$ \exp \left(x^{2}\right) $$

A famous sequence \(\left\\{f_{n}\right\\}\), called the Fibonacci Sequence after Leonardo Fibonacci, who introduced it around A.D. 1200 , is defined by the recursion formula $$ f_{1}=f_{2}=1, \quad f_{n+2}=f_{n+1}+f_{n} $$ (a) Find \(f_{3}\) through \(f_{10}\) - (b) Let \(\phi=\frac{1}{2}(1+\sqrt{5}) \approx 1.618034\). The Greeks called this number the golden ratio, claiming that a rectangle whose dimensions were in this ratio was "perfect." It can be shown that $$ \begin{aligned} f_{n} &=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right] \\\ &=\frac{1}{\sqrt{5}}\left[\phi^{n}-(-1)^{n} \phi^{-n}\right] \end{aligned} $$ Check that this gives the right result for \(n=1\) and \(n=2\). The general result can be proved by induction (it is a nice challenge). More in line with this section, use this explicit formula to prove that \(\lim _{n \rightarrow \infty} f_{n+1} / f_{n}=\phi\). (c) Using the limit just proved, show that \(\phi\) satisfies the equation \(x^{2}-x-1=0\). Then, in another interesting twist, use the Quadratic Formula to show that the two roots of this equation are \(\phi\) and \(-1 / \phi\), two numbers that occur in the explicit formula for \(f_{n}\).

Let \(S=\left\\{x: x\right.\) is rational and \(\left.x^{2}<2\right\\}\). Convince yourself that \(S\) does not have a least upper bound in the rational numbers, but does have such a bound in the real numbers. In other words, the sequence of rational numbers \(1,1.4,1.41,1.414, \ldots\), has no limit within the rational numbers.

In Problems 54-59, use the fact that \(\lim _{x \rightarrow \infty} f(x)=\lim _{x \rightarrow 0^{+}} f\left(\frac{1}{x}\right)\) to find the limits. \(\lim _{n \rightarrow \infty}\left(\frac{n-1}{n+1}\right)^{n}\)

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