Basic maths-calculus-Q13

+1 vote

lim_{n\rightarrow \infty} \frac{1+2+3+..+n}{n^2}, n \in N, is equal to ?

(A). 0

(B). 1

(C). 1/2

(D). 1/4

asked May 19 in Basic Maths by gbeditor (44,560 points)
recategorized Jun 9 by deepak-gatebook

1 Answer

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Best answer

We know that 1+2+3+ \dots n = n(n+1)/2

lim_{n\rightarrow \infty} \frac{1+2+3+..+n}{n^2}

lim_{n\rightarrow \infty} \frac{n(n+1)/2}{n^2}

lim_{n\rightarrow \infty} \frac{n^2+n}{2n^2}

lim_{n\rightarrow \infty} \frac{1+(1/n)}{2}

= 1/2

answered May 27 by deepak-gatebook (112,760 points)
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