Question

If the nth partial sum of summation asubn from n=1 to infinity is equal to

(4n-1)/(n+1)
what is the sum of the summation?

Answer 1

take the limit as n goes to infinity.

Answer 2

wouldn't that mean that it diverges? The answers are whole numbers

Answer 3

You have:\[\sum_{n=1}^\infty a_n\]and you are told:\[\sum_{n=1}^{N}a_n=\frac{4N-1}{N+1}\] Those are the partial sums. Since:\[\lim_{N\rightarrow \infty} \sum _{n=1}^{N}a_n =\sum_{n=1}^{\infty}a_n\] to figure out the sum of the infinite series, take the limit of the partial sum as N goes to infinity.

Answer 4

Thank you, I think I was just confused about the idea that the limit could be equal to the sum.