Limit regarding k = 1 n k a k \sum\limits_{k=1}^n ka_k

Calculus Level 3

The sequence { a n } \{a_n\} satisfy lim n a n = 2020 \lim \limits_{n \to \infty} a_n = 2020 . Find

lim n 1 n 2 k = 1 n k a k \lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^n ka_k


The answer is 1010.

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1 solution

Pi Han Goh
Oct 17, 2020

Apply Stolz–Cesàro theorem ! Let a n = k = 1 n k a k a_n = \sum \limits_{k=1}^n k a_k and b n = n 2 b_n = n^2 .

The limit in question, lim n a n b n \lim\limits_{n\to\infty} \frac{a_n}{b_n} exists, if and only if the limit below exist. And if so, they are equal to each other.

lim n a n + 1 a n b n + 1 b n \lim\limits_{n\to\infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n}

We have lim n a n + 1 a n b n + 1 b n = lim n ( n + 1 ) a n 2 n + 1 = lim n ( 1 + 1 / n ) a n 2 + 1 / n = ( 1 + 0 ) 2020 2 + 0 = 1010 . \lim\limits_{n\to\infty} \frac{a_{n+1} - a_n}{b_{n+1} - b_n} =\lim\limits_{n\to\infty} \dfrac{(n+1)a_n}{2n + 1} =\lim\limits_{n\to\infty} \dfrac{(1 + 1/n) \cdot a_n}{2 + 1/n} = \dfrac{ (1 + 0) \cdot 2020}{2 + 0} = \boxed{1010}.

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