Find the limit: lim_{n->infty} (1/n) sum{k=1,n} cos[(4kpi)/(9n)] (show me all steps)

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jm15

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hi , I really need help for this. how do I evaluate the n inside the cos function and why is there a 1/n next to the sigma notation? can anyone please do a step by step calculation?
 
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jm15 said:
View attachment 34867

hi , I really need help for this. how do I evaluate the n inside the cos function and why is there a 1/n next to the sigma notation? can anyone please do a step by step calculation?
Click to expand...
The question doesn't want you to evaluate the limit of the Riemann Sum. It wants to you convert the expression into the definite integral and evaluate it. Recall that
[math]\lim_{n \to \infty} \dfrac{b-a}{n}\sum_{k=1}^{n}f\left(a + \dfrac{b-a}{n}\cdot k\right) = \int_{a}^{b}f(t)~dt[/math]
They gave you [imath]a=0, b = \frac{4\pi}{9}[/imath]. Can you figure out what's [imath]f(t)?[/imath]
 
My teacher told me earlier to match the above function with the the right-endpoint Riemann sum notation. I actually found cosx in the end. Then I used FTC 2 and found sin(4pi/9) in the end. When I plug that in the answer box, it’s wrong though.
 
Last edited:
jm15 said:
My teacher told me earlier to match the above function with the the right-endpoint Riemann sum notation. I actually found cosx in the end. Then I used FTC 2 and found sin(4pi/9) in the end. When I plug that in the answer box, it’s wrong though.
Click to expand...
Can you show us exactly what you entered, and what the program showed when you did so? It may be just a matter of using the wrong format, or something more subtle.
 
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My final solution is actually sin(4pi/9)-sin(0) not cos(4pi/9)-cos(0)
 
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