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Limit of a sequence

Let f:\mathbb{R}\rightarrow \mathbb{R} be a function such that f(0)=0 and f is differentiable at 0. Let us set

    \[a_n =\sum_{k=1}^{n} f \left( \frac{k}{n^2} \right)\]

Evaluate the limit \lim \limits_{n \rightarrow +\infty} a_n.

Solution

Since f is differentiable at 0 , there is some \varepsilon: x \mapsto \varepsilon(x) such that

    \[f(x) = f(0) + x f'(0) + x \varepsilon(x) \quad , \quad \varepsilon(0) =0\]

and \varepsilon is of course continuous.

Thus,

    \[\sum_{k=1}^{n} f\left ( \frac{k}{n^2} \right ) = \frac{f'(0)}{n} \sum_{k=1}^{n} \frac{k}{n} + \sum_{k=1}^{n} \frac{k}{n^2} \varepsilon \left ( \frac{k}{n^2} \right )\]

Let \epsilon>0. There exists \delta>0 such that |x| \leq \delta which in return means that |\varepsilon(x)| \leq \epsilon. Hence , for n larger than \frac{1}{\delta}+1 it holds that

    \[\left| \sum_{k=1}^n \frac{k}{n^2}\varepsilon \left( \frac{k}{n^2} \right) \right|\leq \epsilon \frac {n}{n}  = \epsilon\]

On the other hand , the sum \displaystyle \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} is a Riemann sum and converges to \frac{1}{2}.

In conclusion,

    \[\lim_{n \rightarrow +\infty} a_n = \frac{f'(0)}{2}\]

 

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