/*! 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 45 Find the sum of each series. $... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the sum of each series. $$\sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$

Short Answer

Expert verified
The sum of the series is 1.

Step by step solution

01

Understand the Series Pattern

Look at the series term: \( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \). This is a telescoping series. In a telescoping series, most terms cancel each other out when you add the series together.
02

Explore the First Few Terms

Write out the first few terms to observe the cancellation pattern:\[\left( \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} \right) + \left( \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} \right) + \left( \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{4}} \right) + \ldots\]Notice that except for the very first term \( \frac{1}{\sqrt{1}} \), each \( -\frac{1}{\sqrt{n+1}} \) cancels with the \( \frac{1}{\sqrt{n}} \) of the next term.
03

Identify the Remaining Terms When Summed Up

Due to the cancellation of middle terms in a telescoping series, all terms cancel out except the first and last terms of the series. Since the series is infinite, the last term approaches zero as \( n \to \infty \).
04

Conclusion on the Sum of the Series

Since all terms cancel out except the very first term, the infinite series converges to the first term of the series. That is, the sum of the series is simply the value of the first term, \( \frac{1}{\sqrt{1}} = 1 \).

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Key Concepts

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

Infinite Series
An infinite series is the sum of terms that extend indefinitely. Imagine adding numbers forever. This is one way to think about infinite series. In mathematics, an infinite series is typically represented by a summation notation \[ \sum_{n=1}^{\infty} a_n \] which indicates adding up terms from the first term to infinity. Infinity isn't a number but a concept representing something without bounds. It's crucial to understand how mathematicians handle such series without going 'to infinity and beyond'. By studying series, we're diving deep into patterns and long-term behavior of sequences.Despite its expansive nature, not all series grow endlessly in value. Instead, some series approach a specific limit, a concept known as convergence. Observing patterns in these series can reveal whether they settle to definite sums or diverge wildly.
Convergence
Convergence is an essential idea when dealing with infinite series. It tells us whether an infinite series settles down to a specific number as more terms are added. When each step brings us closer to a particular sum, the series converges. If this doesn't happen and the series increases indefinitely, it diverges.For a series to converge, the effects of adding more terms should gradually diminish. Consider the series \[ \sum_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right). \] We identify that every additional pair of terms doesn't contribute significantly after simplification. It's a telescoping series, where sequential cancellations reduce the influence of later terms.In such cases, even though the series might involve an infinite number of operations, it converges to a finite value. The sum of the series ends up being just the first few leftover terms, here the very first, which is 1.
Series Pattern Recognition
Pattern recognition in series is a skill that makes understanding them far simpler. In a telescoping series like \[ \sum_{n=1}^{\infty} \left( \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}} \right), \] identifying the cancellation pattern is key. Writing out a few terms as:
  • \( \frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + \cdots \)
reveals that each term typically cancels out with part of the next.Recognizing this telescoping nature is like finding a secret shortcut through a complicated puzzle. Instead of handling each addition individually, most terms "collapse" to zero, leaving only the significant terms unearthed at the start and end. With telescoping series, the magic lies in seeing which terms remain when the dust settles.By systematically reviewing the pattern, students learn which series can be simplified and where the direct computation is needed. Such insights are powerful, saving both time and effort.

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Most popular questions from this chapter

In the alternating harmonic series, suppose the goal is to arrange the terms to get a new series that converges to \(-1 / 2 .\) Start the new arrangement with the first negative term, which is \(-1 / 2\). Whenever you have a sum that is less than or equal to \(-1 / 2,\) start introducing positive terms, taken in order, until the new total is greater than \(-1 / 2 .\) Then add negative terms until the total is less than or equal to \(-1 / 2\) again. Continue this process until your partial sums have been above the target at least three times and finish at or below it. If \(s_{n}\) is the sum of the first \(n\) terms of your new series, plot the points \(\left(n, s_{n}\right)\) to illustrate how the sums are behaving.

a. Show that the improper integral $$ \int_{2}^{\infty} \frac{d x}{x(\ln x)^{p}} \quad(p \text { a positive constant }) $$ converges if and only if \(p>1\) b. What implications does the fact in part (a) have for the convergence of the series $$ \sum_{n=2}^{\infty} \frac{1}{n(\ln n)^{p}} ? $$ Give reasons for your answer.

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 \(\left|a_{n}-L\right| \leq 0.01\) for \(n \geq N .\) How far in the sequence do you have to get for the terms to lie within 0.0001 of \(L ?\) \(a_{1}=1, \quad a_{n+1}=a_{n}+\frac{1}{5^{n}}\)

a. Series for \(\sinh ^{-1} x \quad\) Find the first four nonzero terms of the Taylor series for $$ \sinh ^{-1} x=\int_{0}^{x} \frac{d t}{\sqrt{1+t^{2}}} $$ b. Use the first three terms of the series in part (a) to estimate \(\sinh ^{-1} 0.25 .\) Give an upper bound for the magnitude of the estimation error.

Prove that a sequence \(\left\\{a_{n}\right\\}\) converges to 0 if and only if the sequence of absolute values \(\left\\{\left|a_{n}\right|\right\\}\) converges to 0 .

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