Classic Limits

Calculus Level 3

If S n = ∑ k = 1 n a k \displaystyle S_n = \sum_{k=1}^n a_k and lim ⁡ n → ∞ a n = a \displaystyle \lim_{n\to\infty} a_n = a , then simplify lim ⁡ n → ∞ S n + 1 − S n ∑ k = 1 n k \displaystyle \lim_{n\to\infty} \dfrac{S_{n+1} - S_n}{\sqrt{\sum_{k=1}^n k}} .

a a 2 a \sqrt2 a 0 0 2 a 2a

Moulik Bhattacharya by Moulik Bhattacharya , 20, India

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1 solution

Rewrite

S n + 1 − S n = ∑ k = 1 n + 1 a k − ∑ k = 1 n a k = a n + 1 + ∑ k = 1 n a k − ∑ k = 1 n a k = a n + 1 , S_{n+1} - S_{n} = \sum_{k=1}^{n+1} a_k - \sum_{k=1}^{n} a_k = a_{n+1} + \sum_{k=1}^{n} a_k - \sum_{k=1}^{n} a_k = a_{n+1},

and

∑ k = 1 n k = n ( n + 1 ) 2 . \sum_{k=1}^{n} k = \frac{n(n+1)}{2}.

The limit we want to find L L then becomes

L = 2 lim ⁡ n → ∞ a n + 1 ( n + 1 ) n = 0 , L = \sqrt{2} \lim_{n\to \infty} \frac{a_{n+1}}{\sqrt{(n+1)n}} = 0,

since the sequence ( a n ) 0 ∞ \left( a_n \right)_0^{\infty} is limited by a a and the sequence ( ( n + 1 ) n ) 0 ∞ \left( (n+1)n \right)_0^{\infty} diverges as n → ∞ n \rightarrow \infty .

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