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  4. displaystyle limn → ∞(1+2+3+...+n/n2), n ∈ N, is equal to

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Q. $\displaystyle \lim_{n \to \infty}$$\frac{1+2+3+...+n}{n^{2}}$, $n \in N$, is equal to

Limits and Derivatives

Solution:

As $\displaystyle \lim_{n \to \infty}$$\frac{1+2+3+...+n}{n^{2}}$
$=\displaystyle \lim_{n \to \infty}$$\frac{n\left(n+1\right)}{2n^{2}}$
$=\displaystyle \lim_{n \to \infty}$$\frac{1}{2}\left(1+\frac{1}{n}\right)$
$=\frac{1}{2}$