distribution of limits for infinite terms

18 Views Asked by At

enter image description here Please tell me which one is correct answer and how because I m not getting it. If I apply limit simultaneously I should get 0 but I have confusion.

1

There are 1 best solutions below

3
On BEST ANSWER

You have:

$\lim_\limits{n\to\infty}\left(\frac {f(a + \frac {1}{n^2}) - f(a)}{\frac {1}{n^2}} + 2\frac {f(a + \frac {2}{n^2}) - f(a)}{\frac {2}{n^2}} + 3\frac {f(a + \frac {3}{n^2}) - f(a)}{\frac {3}{n^2}}+\cdots + n\frac {f(a + \frac {n}{n^2}) - f(a)}{\frac {n}{n^2}}\right)\frac{1}{n^2}$

For each term:

$\lim_\limits{n\to\infty}\frac {f(a + \frac {k}{n^2}) - f(a)}{\frac k{n^2}} = f'(a)$

$\lim_\limits{n\to\infty}\left(f'(a) + 2 f'(a) + 3 f'(a) + \cdots nf'(a)\right)\frac{1}{n^2}$

Which is an arithmetic series.

$\lim_\limits{n\to\infty} \frac{n(n+1)f'(a)}{2n^2}\\ \frac 12 f'(a)$

Update reflecting a learner's comment.

If we want to do it "right" we need to start with the Taylor expansion.

$f(a+h) = f(a) + hf'(a) + O(h^2)$

$\lim_\limits{n\to\infty} \sum_\limits{k=1}^n f(a+\frac {k}{n^2}) - nf(a)\\ \lim_\limits{n\to\infty} \sum_\limits{k=1}^n \left(f(a) + \frac {k}{n^2}f'(a) + k^2 O(\frac 1{n^4})\right) - nf(a)\\ \lim_\limits{n\to\infty} nf(a) + \frac {(n)(n+1)}{2n^2}f'(a) + \frac {n(n+1)(2n+1)}{6} O(\frac 1{n^4}) - nf(a)\\ \lim_\limits{n\to\infty} \frac {(n)(n+1)}{2n^2}f'(a) + O(n^{-1})\\ \frac 12 f'(a)$