Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
portion

portion

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter10: Sequences, Series, And Probability
Section10.2: Arithmetic Sequences
Problem 68E
See similar textbooks
icon
Related questions
Question

I've looked at the existing answer to this question already but it doesn't quite make sense to me.
First, when I plug in a random positive integer for N, the formula given as the answer to this problem doesn't produce the same outcome as when I plug the summation into an online calculator. 

Second, I'm having a hard time understanding how we can limit the answer formula to a set number of parts for all possible positive  integers of N. Is it not that for all N, the inner portion of the by parts equation 1/N[f(n)-f(n+N)] when put into summation would have all values from f(1) to f(N) and also have N number of remaining portions at the end uncanceled? How can this be simplified to work for any N? 

I think I have a difficult time reasoning with infinity so perhaps that's where the issue lies. Any help would be greatly appreciated, Thank You!

To determine To find: 1 a formula for > n(n + N) n=1 Answer to Problem 114E The formula for 00 Σ ☆(1+ 글 + N+ ). n(n + N) n=1 Explanation of Solution Given: 00 1 Σ n(n + N) n=1 N is a positive integer Calculation: 00 1 Consider the series > п(п + N) n=1 00 00 1 n+N) f(n) = f(n + N) = п(n + N) n=1 n=1 let an of(n – 1) = || n(n+N) (n+N) (п-1) → an = f(n) – f(n+ N)) for n=1 = S1 = aj = 7F(1) – f(1+ N)) for n=2 = S2 = a¡ + a2 *F(1) + f(2) – f(1 + N) – f(2 + N)) for n=3 = S3 = a1 + a2 + a3 = 7f(1) +f(2) +f(3) – f(1 + N) – f(2+ N) - f(3 + N) n → 0 → f(n) → ∞ then S„=¬f(1) + f(2) + S„=#(1+ +N +) - f(N – 1)) Therefore, 00 1 (1 + +N + ). n(n + N) n=1
Transcribed Image Text:To determine To find: 1 a formula for > n(n + N) n=1 Answer to Problem 114E The formula for 00 Σ ☆(1+ 글 + N+ ). n(n + N) n=1 Explanation of Solution Given: 00 1 Σ n(n + N) n=1 N is a positive integer Calculation: 00 1 Consider the series > п(п + N) n=1 00 00 1 n+N) f(n) = f(n + N) = п(n + N) n=1 n=1 let an of(n – 1) = || n(n+N) (n+N) (п-1) → an = f(n) – f(n+ N)) for n=1 = S1 = aj = 7F(1) – f(1+ N)) for n=2 = S2 = a¡ + a2 *F(1) + f(2) – f(1 + N) – f(2 + N)) for n=3 = S3 = a1 + a2 + a3 = 7f(1) +f(2) +f(3) – f(1 + N) – f(2+ N) - f(3 + N) n → 0 → f(n) → ∞ then S„=¬f(1) + f(2) + S„=#(1+ +N +) - f(N – 1)) Therefore, 00 1 (1 + +N + ). n(n + N) n=1
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
College Algebra
College Algebra
Algebra
ISBN:
9781938168383
Author:
Jay Abramson
Publisher:
OpenStax
SEE MORE TEXTBOOKS