Limit is Sum

jeffer vitola

New member
hello ........... I propose this exercise for you to solve on various methods .......

$\lim_{n \to{+}\infty}{\frac{1}{n}\sum_{i=1}^n({1+\frac{i}{n}}})^{-2}$

thanks

att
jefferson alexander vitola

MarkFL

Staff member
I have moved this topic, as it seems to be posted as a challenge rather than for help.

ZaidAlyafey

Well-known member
MHB Math Helper
$\lim_{n \to{+}\infty}{\frac{1}{n}\sum_{i=1}^n({1+\frac{i}{n}}})^{-2}$
This can be solved directly using the Riemann sum

$$\displaystyle \lim_{n \to{+}\infty} \sum_{i=1}^n \frac{n}{(n+i)^2}=\int^2_1 \frac{1}{x^2}\, dx = \frac{1}{2}$$

jeffer vitola

New member
This can be solved directly using the Riemann sum

$$\displaystyle \lim_{n \to{+}\infty} \sum_{i=1}^n \frac{n}{(n+i)^2}=\int^2_1 \frac{1}{x^2}\, dx = \frac{1}{2}$$
hello........interesting, but as I said in my previous forum topic or main focus is that you develop by various methods ...... one can be evaluated by the summation properties and then calculate its limit for example,,,,,,,, if you are cant make exercise , there is not problem,,,, thanks........

att
jefferson alexander vitola

MarkFL

Staff member
hello........interesting, but as I said in my previous forum topic or main focus is that you develop by various methods ...... one can be evaluated by the summation properties and then calculate its limit for example,,,,,,,, if you are cant make exercise , there is not problem,,,, thanks........

att
jefferson alexander vitola
Zaid has shown you a very straightforward method to evaluate the sum. Your initial post said only "solve on various methods." And this is what Zaid has done.

Why don't you demonstrate the technique you have? We expect that when people post problems as a challenge, they have a solution which they post if the problem has not been solved within about a week's time. Although this problem has been solved, but seemingly not to your satisfaction, it is now time for you to show us your solution.