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)
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)
\(a_n=\frac{1}{n^3+1}+\frac{2^2}{n^3+2}+...+\frac{n^2}{n^3+n}\)
then the sequence \(\left\{a_n\right\}_{n=1}^{\infin}\) converges to ____________ (rounded off to two decimal places)
Updated On: Oct 1, 2024
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Correct Answer: 0.3
Solution and Explanation
The correct answer is 0.30 to 0.40. (approx)
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