I have a question about the limit of Riemann’s sum. My Riemann’s sum is $\displaystyle \sum_{i=1}^{2n} \left(3 + \left(\frac{i}{n}\right)^2\right)\frac{1}{n}$. I tried to calculate the values of this sum using Sage but it was increasing as n was 10, 100, 1000, and did not approach any number. However, Symbolab says that there is a limit actually and it is 26/3.
Can you please help me to write integral and solve it for this problem?
Thank you a lot!
On
Hint: Note that \begin{eqnarray*} \sum_{i=1}^{2n} \left(3 + \left(\frac{i}{n}\right)^2\right)\frac{1}{n}&=&\sum_{1\leq i \leq 2n} \frac{3+\left(\frac{i}{n}\right)}{n}\\ &=&\sum_{1\leq i \leq 2n}\frac{3}{n}+\frac{i^{2}}{nn^{2}}\\ &=&\sum_{1\leq i \leq 2n}\frac{3}{n}+n^{-3}i^{2}\\ &=&\frac{1}{3n^{2}}+\frac{2}{n}+\frac{26}{3}. \end{eqnarray*} This's a hint, so I think you can complete each the details.
\begin{gather*} \frac{1}{n}\sum ^{2n}_{i=1}\left( 3+\frac{i^{2}}{n^{2}}\right) =\frac{1}{n}\left( 6n+\frac{2n( 2n+1)( 4n+1)}{6n^{2}}\right)\\ \lim _{n\rightarrow \infty }\frac{1}{n}\left( 6n+\frac{2n( 2n+1)( 4n+1)}{6n^{2}}\right) =\frac{26}{3}\\ \\ \end{gather*}