/*! 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 65 $\lim _{n \rightarrow \infty} \f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 65

$\lim _{n \rightarrow \infty} \frac{1 . n+(n-1)(1+2)+(n-2)(1+2+3)+. .1 \cdot \sum_{r=1}^{n} r}{n^{4}}$ is equal to (A) \(1 / 12\) (B) \(1 / 24\) (C) \(1 / 6\) (D) \(1 / 48\)

Short Answer

Expert verified
Answer: The value of the limit is 0.

Step by step solution

01

Identify the Sums in the Expression

First, we can rewrite the expression as follows: \(\lim _{n \rightarrow \infty} \frac{n+ (n-1)\sum_{r=1}^{2} r + (n-2)\sum_{r=1}^{3} r+\cdots+1\cdot\sum_{r=1}^{n} r}{n^{4}}\) Now we can focus on finding the sum of the series in the expression.
02

Compute the Sum for Each Term

The sum of the first 'k' natural numbers can be expressed as: \(\sum_{r=1}^{k} r = \frac{k(k+1)}{2}\) So, our expression becomes: \(\lim _{n \rightarrow \infty} \frac{n+ (n-1)\frac{3}{2} + (n-2)\frac{6}{2}+\cdots+(1)\frac{n(n+1)}{2}}{n^{4}}\)
03

Factor Out the Common Factor

Now, let's factor out \(\frac{1}{2}\) from each term in the numerator: \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}(2n+ (n-1)(3) + (n-2)(6)+\cdots+(1)(n(n+1)))}{n^{4}}\)
04

Simplify the Expression

Now, let's expand the series in the expression: \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}(2n+ 3n-3 + 6n-12+\cdots+n^{2}+n)}{n^{4}}\) Next, we add the terms: \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}((n+1)(n+2)-3n)}{n^{4}}\)
05

Compute the Limit

Now, divide each term by \(n^{3}\) and find the limit: \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}(\frac{(n+1)(n+2)}{n^3} - \frac{3n}{n^3})}{n}\) As \(n \rightarrow \infty\), \(\frac{(n+1)(n+2)}{n^3} \rightarrow 1\) and \(\frac{3n}{n^3} \rightarrow 0\): \(\lim _{n \rightarrow \infty} \frac{\frac{1}{2}(1- 0)}{n}=0\)
06

Check the Options

Since the value of the limit is 0, none of the options are correct. To double-check this result, you may want to review the problem statement and solution, or perhaps consult your teacher for any errors or mistakes.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

$\lim _{x \rightarrow 0} \frac{1}{x}\left(\sqrt{\frac{1}{x^{2}}+1}-\frac{1}{x}\right)+x \ln \left(1+a^{1 / x}\right), a>0, a \neq$ (A) a (B) (C) \(1+\mathrm{a}\) (D) None of these

$\lim _{x \rightarrow 0} \frac{\int_{\sqrt{x}}^{x^{2}} \tan ^{-1}\left(\frac{t^{2}}{1+t^{2}}\right) \text { dt }}{\sin 2 x}$ is equal to (A) 0 (B) 1 (C) \(1 / 2\) (D) does not exist

$\lim _{x \rightarrow-\infty}\left\\{x+\sqrt{x^{2}+3 x \cos \frac{1}{|x|}}\right\\}$ is equal to (A) \(3 / 2\) (B) \(-3 / 2\) (C) \(-1\) (D) none of these

$\lim _{\mathrm{n} \rightarrow \infty}\left(\frac{\mathrm{n}}{\mathrm{n}^{2}-2}+\frac{4^{\mathrm{n}}(-1)^{\mathrm{n}}}{2^{\mathrm{n}}-1}\right)^{-1}$ is equal to (A) 2 (B) (C) 0 (D) None of these

If $\lim _{x \rightarrow \infty} \frac{729^{x}-243^{x}-81^{x}+9^{x}+3^{x}-1}{x^{3}}=k(\ln 3)^{3}\(, then \)k$ is equal to (A) 4 (B) 5 (C) 6 (D) none
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

Read Explanation

Applied Mathematics

Read Explanation

Calculus

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.