Question

For n ∈ \(\N\), if
\(a_n=\frac{1}{n^3+1}+\frac{2^2}{n^3+2}+...+\frac{n^2}{n^3+n}\)
then the sequence \(\left\{a_n\right\}_{n=1}^{\infin}\) converges to ____________ (rounded off to two decimal places)

Answer 1

Correct Answer: 0.3

The given sequence involves the sum of fractions that behave like \( \frac{k^2}{n^3 + k} \) as \( n \to \infty \). For large \( n \), each term behaves like \( \frac{k^2}{n^3} \), and the sum of such terms converges to a value approximately 0.30 when evaluated. Thus, the correct answer is 0.30.