|
Title: Compute the sum and the limit of the sum as n -> infinity. [tex]{\sum_{n=1}^n \ Post by: quixie on May 7, 2015 Compute the sum and the limit of the sum as n -> infinity.
\({\sum_{n=1}^n \frac{1}{n}[13(i/n)^2-9(i/n)]}\) a. Sum = \(\frac{13(n+1)(2n+1)}{6n^2}-\frac{9(n+1)}{2n}\) ; limit of sum as n -> infinity is -1/6 b. Sum = \(\frac{(n+1)(2n+1)}{13n^2}-\frac{n+1}{9n}\) ; limit of sum as n -> infinity is -17/117 c. Sum = \(\frac{13(n+1)(2n+1)}{6n^2}-\frac{9(n+1)}{2n}\) ; limit of sum as n -> infinity is 0 d. Sum = \(\frac{(n+1)(2n+1)}{13n^2}-\frac{n+1}{9n}\) ; limit of sum as n -> infinity is 17/117 |