Simple calculus question
Name:
Anonymous
2008-11-13 9:45
Evaluate
\lim_{n\to\infty} \frac{1}{n} \sum^{n-1}_{k=0} (n-k) \frac{dx}{x^4 + 1}.
Name:
Anonymous
2008-11-13 9:47
Oops, I mean:
\lim_{n\to\infty} \frac{1}{n} \sum^{n-1}_{k=0} (n-k) \int^{\frac{k+1}{n}}_{\frac{k}{n}} \frac{dx}{x^4 + 1}
Name:
Anonymous
2008-11-13 11:44
Still don't see it.
Name:
Anonymous
2008-11-13 12:26
Doesn't look so simple from where I'm standing.
Name:
Anonymous
2008-11-13 13:21
I CANT DO IT.
Name:
Anonymous
2008-11-13 16:08
Evaluate the integral first.
Name:
4tran
2008-11-13 18:53
\lim_{n\to\infty} \sum^{n-1}_{k=0} (1-\frac{k}{n}) \int^{\frac{k+1}{n}}_{\frac{k}{n}} \frac{dx}{x^4 + 1} =
\lim_{n\to\infty} [\sum^{n-1}_{k=0} \int^{\frac{k+1}{n}}_{\frac{k}{n}} \frac{dx}{x^4 + 1}] - [\sum^{n-1}_{k=0} \frac{k}{n} \int^{\frac{k+1}{n}}_{\frac{k}{n}} \frac{dx}{x^4 + 1}]
The 1st term should be obvious, and you should be able to figure out what the 2nd term means.
Name:
Anonymous
2008-11-14 0:09
Telescoping?
Name:
Anonymous
2008-11-14 2:50
\lim_{n\to\infty} \frac{1}{n} \sum^{n-1}_{k=0} f(c) (1 - \frac{k}{n}) = \int^1_0 f(c)(1-x) dx
Name:
Anonymous
2008-11-14 17:07
well the limit of 1/n as n nears infinity is zero.
zero times the rest of that is equal to zero.
this answer is zero.
or lrn2parenthesis.
Name:
Anonymous
2008-11-15 21:35
>>10
Wrong. Get an education.