If S n - k -1 n a k and lim n - a n - a- then lim n - S n -1- S n - k -1 n k is equal toA- 0B- a C- -2 aD- 2 a

The correct option is A 0n→∞limS_n+1 S_n/√(Σ _k=1^nk)=n→∞lima_n+1/√(n(n+1)/2)=0 (∵ a_n+1=a) ...

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Standard Limits to Remove Indeterminate Form

If S n =Σ k =...

Question

If $S_{n} = Σ_{k = 1}^{n} a_{k} a n d l i m n \to \infty a_{n} = a, t h e n l i m n \to \infty \frac{S_{n + 1} - S_{n}}{\sqrt{Σ_{k = 1}^{n} k}}$ is equal to

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S_{n} = n \sum k = 1 a_{n}

lim n \to \infty a_{n} = a

lim n \to \infty \frac{S_{n + 1} - S_{n}}{\sqrt{\sum_{1}^{n} k}} =

S_{n} = n \sum k = 1 \frac{k}{(k - 1)^{4 / 3} + (k^{2} - 1)^{2 / 3} + (k + 1)^{4 / 3}}

lim n \to \infty \frac{S_{n}}{n^{2 / 3}} = \frac{1}{p} .

p

S_{n} = [\frac{1}{1 + \sqrt{n}} + \frac{1}{2 + \sqrt{n}} + . . . + \frac{1}{n + \sqrt{n^{2}}}]

lim n \to \infty S_{n}

S_{n} = \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . . .

n

lim n \to \infty S_{n}

s_{n} = \frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . .

n

lim n \to \infty s_{n}

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