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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
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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
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