limn→∞([itex]\frac{1}{n+1}[/itex]+[itex]\frac{1}{n+2}[/itex]+...+[itex]\frac{1}{4n}[/itex])
Hi, when I first looked at this limit I thought that the solution is 0, but the assistant applied Riemann Sum and she found that this limit equals to ln4. Why this limit is not 0. I'm confused. Can you help me?
Hi, when I first looked at this limit I thought that the solution is 0, but the assistant applied Riemann Sum and she found that this limit equals to ln4. Why this limit is not 0. I'm confused. Can you help me?
via Physics Forums RSS Feed http://www.physicsforums.com/showthread.php?t=725797&goto=newpost
0 commentaires:
Enregistrer un commentaire