I have calculus midterm in exactly 1 hour, please help answer this question

[TGCid]

I have a calculus midterm in an hour. The professor gave us some pratice problems to do. I did them all fine except for one. I don't understand #6 on http://www.math.ucsd.edu/~dlittle/courses/2001/20B/practice1.pdf
It would be really great if someone would explain to me how to solve it. COME ON GUYS!

 

[DAM]

the lim as n-> inf = 0, thats the answer.





dam()

 

[khtm]

<< It would be really great if someone would explain to me how to solve it. >>



Good job, DAM.

 

[DAM]

very well then,


I did not do the calculus that is required however if you just look at the problem, the summation is reaching 1/inf which equals to zero. just by inpection you get the answer.


btw, I took calc 3 like 2 years ago so im rusty as hell.


dam()

 

[khtm]

My calculus is getting old too, but I'll try (this won't be a mathematically correct answer, though):

The original question changes to -> (integral from 1 to n)[1/(sqrt(n)sqrt(n+x))] where x can be any variable. This is now a definite integral, as the question requested. By solving this integral, we get [1/sqrt(n)][(1/sqrt(2n)) - (1/sqrt(n+1)). Now if you take the limit as n-> infinity, you can see that all terms will go to 0 as n-> infinity.

Sorry if that was a bad answer

 

[raptor13]

Divide the entire thing by n. You'll end up with (1/n) in the numerator and the square root of one times the square root of one on the bottom. The denominator is a constant one and the numerator is approaching zero and n approches infinity, hench the whole series approches zero.

 

[TGCid]

I took the midterm. I did fairly decent. Luckily that type of question wasn't on it. Thanks guys.