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Evaluate the following telescoping series or state whether the series diverges. $$ \sum_{n=1}^{\infty}(\sqrt{n}-\sqrt{n+1}) $$

Short Answer

Expert verified
The series diverges.

Step by step solution

01

Identify the Series Form

The given series is \( \sum_{n=1}^{\infty} (\sqrt{n} - \sqrt{n+1}) \). Notice how it has the form of a telescoping series, where each term consists of two parts: \( \sqrt{n} \) and \(- \sqrt{n+1} \).
02

Write Down Partial Sums

To find the partial sums, calculate \( S_k = \sum_{n=1}^{k} (\sqrt{n} - \sqrt{n+1}) \). When you expand it, most terms will cancel out, except for the first and the last terms: \( S_k = (\sqrt{1} - \sqrt{k+1}) \).
03

Evaluate the Limit of Partial Sums

Calculate the limit of \( S_k \) as \( k \rightarrow \infty \): \( \lim_{k \to \infty} (\sqrt{1} - \sqrt{k+1}) = 1 - \lim_{k \to \infty} \sqrt{k+1} \). The term \( \sqrt{k+1} \) goes to infinity, meaning \( \lim_{k \to \infty} \sqrt{k+1} = \infty \). Therefore, the limit is \( 1 - \infty = -\infty \).
04

Determine Convergence or Divergence

Since the limit of the partial sum \( S_k \) goes to \(-\infty\) as \( k \rightarrow \infty \), the series does not converge to a finite value and instead diverges.

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Key Concepts

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

Infinite Series
In mathematics, an infinite series is a sum of infinitely many terms. It is often represented as \( \sum_{n=1}^{\infty} a_n \), where each \( a_n \) is a term in the series. Infinite series can have important implications in calculus and analysis, providing ways to express functions, solve problems, and understand convergence and divergence.
Unlike finite series, infinite series can often result in surprising behaviors—instead of simply adding up to a larger number, they can either converge to a fixed value or diverge. Working with infinite series helps develop an understanding of limits and the behavior of functions as they extend infinitely. A good grasp of infinite series is fundamental when dealing with functions like exponential, logarithmic, or trigonometric functions through techniques such as Taylor series.
Partial Sums
Partial sums are a key concept in understanding infinite series. A partial sum is the sum of the first \( n \) terms of a series, usually denoted as \( S_n \). For instance, if you have a series \( \sum_{n=1}^{\infty} a_n \), the partial sum for the first \( k \) terms is given by \( S_k = a_1 + a_2 + \cdots + a_k \).
Telescoping series benefit greatly from partial sums because they notably simplify the calculation. In telescoping series, most terms cancel with a corresponding negative from previous or subsequent terms, leaving only a few terms to actually compute. This simplification is essential when dealing with series where direct calculation would otherwise be challenging.
Studying partial sums allows mathematicians to determine whether long, seemingly complex series expressions simplify to manageable forms, often through cancellation.
Series Divergence
Divergence of a series occurs when the partial sums do not converge to a fixed number as the number of terms tends to infinity. In other words, if \( \lim_{k \to \infty} S_k \) does not exist or equals infinity, the series diverges.
A diverging series like \( \sum_{n=1}^{\infty} (\sqrt{n} - \sqrt{n+1}) \) might not settle to a constant value, meaning that as more terms are added, they might push the total sum towards infinity or an undefined form. This divergence is identified by evaluating the limit of the partial sums. Series in mathematics can display several types of behavior based on their convergence or divergence, deeply affecting their application and interpretation in various mathematical contexts.
Limit Evaluation
Limit evaluation involves finding the value that a sequence or function approaches as the index or input values extend towards infinity. It is a crucial component in determining the behavior of series and sequences within calculus.
For series, limit evaluation is used to assess whether the partial sums \( S_k \) converge to a particular number or diverge. If \( \lim_{k \to \infty} S_k \) equals a finite number, the series converges. However, if the limit does not exist or takes an undefined form, the series diverges.
In telescoping series, limit evaluation is effectively streamlined since many series terms offset each other. Nevertheless, understanding the growth or reduction of remaining terms ultimately underscores the series' convergence or divergence. Evaluating limits, therefore, acts as a gateway to deeper insights into series and their implications in both pure and applied mathematics.

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Most popular questions from this chapter

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}<1 / 2\), then \(\sum a_{n}\) converges, while if \(\lim _{n \rightarrow \infty} \frac{a_{2 n+1}}{a_{n}}>1 / 2\), then \(\sum a_{n}\) diverges. Let \(a_{n}=\frac{1}{4} \frac{3}{6} \frac{5}{8} \cdots \frac{2 n-1}{2 n+2}=\frac{1 \cdot 3-5 \cdots(2 n-1)}{2^{n}(n+1) !} .\) Explain why the ratio test cannot determine convergence of \(\sum_{n=1}^{\infty} a_{n} .\) Use the fact that \(1-1 /(4 k)\) is increasing \(k\) to estimate \(\lim _{n \rightarrow \infty} \frac{a_{2 n}}{a_{n}}\).

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit. $$ a_{n}=\frac{\ln (\cos n)}{n} $$

Use the root test to determine whether \(\sum_{m=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is as follows. $$ a_{k}=\left(\frac{2 k^{2}-1}{k^{2}+3}\right)^{k} $$

The following series do not satisfy the hypotheses of the alternating series test as stated. In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.\(\sum_{n=1}^{\infty}(-1)^{n+1} \frac{\sin ^{2} n}{n}\)

Use the ratio test to determine whether \(\sum_{n=1}^{\infty} a_{n}\) converges, where \(a_{n}\) is given in the following problems. State if the ratio test is inconclusive. $$ \sum_{n=1}^{\infty} \frac{\left(2^{n} n !\right)^{2}}{(2 n)^{2 n}} $$

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