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In the problem we have the inequality: $x - \frac{x^{2}}{2} < ln(1 + x) < x $ for $x > 0$

And we need to use to find : $\lim \Pi_{k=1}^{n} (1 + \frac{k}{n^{2}}) $

Taking $x = \frac{k}{n²}$ and using the function $e^{x}$ I get:

$e^{\sum_{k=1}^{n} \frac{k}{n²} + \frac{k²}{n^4}} < \Pi_{k=1}^{n} (1 + \frac{k}{n^{2}}) < e^{ \sum_{k=1}^{n}\frac{k}{n²}} $

I don't know how calculate the limits $\lim \sum_{k=1}^{n} (\frac{k}{n²} + \frac{k²}{n^4}) $ and $\sum_{k=1}^{n} \frac{k}{n²} $

How can I proceed? Thank you.

Did
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Zouhair El Yaagoubi
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1 Answers1

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$\lim\limits_{n\to\infty}\sum\limits_{k=1}^{n}\frac{k}{n^{2}}=\lim\limits_{n\to\infty}\frac1n\sum\limits_{k=1}^{n}\frac{k}{n}=\int\limits_0^1 xdx=\frac12 $

$\sum\limits_{k=1}^{n}\frac{k^2}{n^{4}}\leq \sum\limits_{k=1}^{n}\frac{n^2}{n^{4}}=\frac1n $

Minz
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