I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.
This is the question:
Consider$\sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $\sum_{k=1}^n 3+\sum_{k=1}^n ( 2/k^3)$
This is the formula we had to use: Click here to see the approximation integral formula
This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ \int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k – (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.
Could someone please tell me how to solve this? Thanks
Best Answer
Hint. You may write $$ \begin{align} \sum_{k=1}^n\left(3+\frac1{k^3} \right)&=\left(3+\frac1{1^3} \right)+\sum_{k=\color{red}{2}}^n\left(3+\frac1{k^3} \right) \\\\&\le \left(3+\frac1{1^3}\right)+\int_{\color{red}{2-1}}^n\left(3+\frac1{x^3} \right)dx \end{align} $$ then getting easily an upper bound.