Pre-knowledge
introduce
According to the concept of definite integral, we can get
∫ a b f ( x ) d x = lim n → + ∞ 1 n ∑ i = 1 n f ( a + b − a n i ) \int_a^bf(x)dx=\lim\limits_{n\to +\infty}\dfrac 1n\sum\limits_{i=1}^{n}f(a+\dfrac{b-a}{n}i) ∫abf(x)dx=n→+∞limn1i=1∑nf(a+nb−ai)
Then, for parts of the form
lim n → + ∞ 1 n ∑ i = 1 n a i \lim\limits_{n\to +\infty}\dfrac 1n\sum\limits_{i=1}^na_i n→+∞limn1i=1∑nai
This kind of formula can be obtained by definite integral.
example
Determine limit n → + ∞ ( nn 2 + 1 + nn 2 + 2 2 + ⋯ + nn 2 + n 2 ) \lim\limits_{n\to+\infty}(\dfrac{n}{n^2+1 }+\dfraction}{n^2+2^2}+\cdots+\dfraction}{n^2+n^2})n→+∞lim(n2+1n+n2+22n+⋯+n2+n2n)
Solution:
\qquad原式 = lim n → + ∞ 1 n ∑ i = 1 n 1 1 + ( i n ) 2 = ∫ 0 1 1 1 + x 2 d x = arctan x ∣ 0 1 = π 4 =\lim\limits_{n\to+\infty}\dfrac 1n\sum\limits_{i=1}^n\dfrac{1}{1+(\frac in)^2}=\int_0^1\dfrac{1}{1+x^2}dx=\arctan x\bigg\vert_0^1=\dfrac{\pi}{4} =n→+∞limn1i=1∑n1+(ni)21=∫011+x21dx=arctanx
01=4p