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#### jeffer vitola

##### New member

- Jun 15, 2013

- 27

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

thanks

att

jefferson alexander vitola

- Thread starter jeffer vitola
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- Thread starter
- #1

- Jun 15, 2013

- 27

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

thanks

att

jefferson alexander vitola

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- Jan 17, 2013

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This can be solved directly using the Riemann sum\[\lim_{n \to{+}\infty}{\frac{1}{n}\sum_{i=1}^n({1+\frac{i}{n}}})^{-2}\]

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

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- Jun 15, 2013

- 27

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........This can be solved directly using the Riemann sum

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

att

jefferson alexander vitola

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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.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

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.