Riemann Sum – Exercise 2330 Post category:Riemann Sum Post comments:0 Comments Exercise Evaluate the limit limn→∞∑k=1n4n2(2k−1)n\lim_{n \rightarrow \infty}\sum_{k=1}^n\frac{4}{n}\sqrt{\frac{2(2k-1)}{n}}n→∞limk=1∑nn4n2(2k−1) Final Answer Show final answer limn→∞∑k=1n4n2(2k−1)n=163\lim_{n \rightarrow \infty}\sum_{k=1}^n\frac{4}{n}\sqrt{\frac{2(2k-1)}{n}}=\frac{16}{3}n→∞limk=1∑nn4n2(2k−1)=316 Solution Coming soon… Share with Friends Read more articles Next PostRiemann Sum – Exercise 2322 You Might Also Like Riemann Sum – Exercise 2311 January 6, 2019 Riemann Sum – Exercise 2318 January 6, 2019 Riemann Sum – Exercise 2322 January 6, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
