JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

y_n=\frac{1^2+2^2+3^2+\cdots+n^2}{n^3}

then

\lim _{n \rightarrow \infty} y_n

is equal to

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

) is correct because the sum of the squares of the first n natural numbers is

\frac{n(n+1)(2n+1)}{6}

, and dividing by

n^3

and taking the limit as

n

approaches infinity yields

\frac{1}{3}

More Questions from LIMITS

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\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

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\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

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f^{\prime}(0)

\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}