2 2 votes What is the correct asymptotic of $\sum^{n}_{i=1} \dfrac {n}{i^2}?$$\Theta(n)$$\Theta(n$ log $n)$$\Theta(1)$$\Theta(n^2)$ Algorithms non-gatecse jest asymptotic-notations + – Random Oracle 467 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
4 4 votes As \(n \to \infty\), the sum \(\sum_{i=1}^n \frac{1}{i^2}\) converges to a constant (proof), so it is \(\Theta(1)\). Hence\[S = n \cdot \Theta(1) = \Theta(n).\](we use asymptotic analysis to understand behavior for large values of \(n\), i.e., as \(n \to \infty\)) Abhishek-1011 answered Jul 8, 2025 Abhishek-1011 comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Option A is Correct soudipta_dutta answered Jul 8, 2025 • edited Feb 11 by soudipta_dutta soudipta_dutta comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes OPTION 'A' IS ANSWER Gopi_Tirumala answered Aug 8, 2025 Gopi_Tirumala comment Share Follow 0 reply Please log in or register to add a comment.