Correct Answer - Option 3 :
\(\frac{1}{2}\)
Formula used:
\(1 + 2 + 3 + ...+ n =\frac{n(n+1)}{2}\)
Calculation:
\(\displaystyle\lim_{n\rightarrow∞}\dfrac{1+2+3+...+n}{n^2}\) = ?
Since we know that \(1 + 2 + 3 + ...+ n =\frac{n(n+1)}{2}\)
⇒ \(\displaystyle\lim_{n\rightarrow∞}\dfrac{1+2+3+...+n}{n^2}\) = \(\displaystyle\lim_{n\rightarrow∞}\frac{\frac{n(n+1)}{2}}{{n^2}}\)
⇒ \(\displaystyle\lim_{n\rightarrow∞}\frac{(n+1)}{{2n}}\)
⇒ \(\displaystyle\lim_{n\rightarrow∞}\frac{n(1+1/n)}{{2n}}\)
⇒ \(\displaystyle\lim_{n\rightarrow∞}\frac{(1+1/n)}{{2}}\)
⇒ \(\frac{1}{2}\) as n → ∞
\(\displaystyle\lim_{n\rightarrow∞}\dfrac{1+2+3+...+n}{n^2} = \frac{1}{2}\), ∀ n ∈ N
Hence, the correct answer is option 3)
