Ta có:
\(\sum\left(n+1\right)n^2=\sum\left(n^2+n^3\right)=\sum n^2+\sum n^3=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}+\dfrac{n^2\left(n+1\right)^2}{4}\)
\(\Rightarrow\lim\dfrac{2.1^2+3.2^2+...+\left(n+1\right)n^2}{n^4}=\lim\dfrac{n\left(n+1\right)\left(2n+1\right)}{6n^4}+\lim\dfrac{n^2\left(n+1\right)^2}{4n^4}\)
\(=\dfrac{2}{6}+\dfrac{1}{4}=\dfrac{7}{12}\)
Tính
lim \(\left(\dfrac{1}{\sqrt{n^3+1}}+\dfrac{1}{\sqrt{n^3+2}}+....+\dfrac{1}{\sqrt{n^3+n}}\right)\)
Tính
lim \(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{n\left(n+1\right)}\right)\)
tính các giới hạn sau:
a) lim (3n2+n2-1)
b)lim \(\dfrac{n^3+3n+1}{2n-n^3}\)
c) lim \(\dfrac{-2n^3+3n+1}{n-n^2}\)
d) lim \(\left(n+\sqrt{n^2-2n}\right)\)
e) lim \(\left(2n-3.2^n+1\right)\)
f) lim \(\left(\sqrt{4n^2-n}-2n\right)\)
g) lim \(\left(\sqrt{n^2+3n-1}-\sqrt[3]{n^3-n}\right)\)
tính các giới hạn sau:
a, lim\(\frac{3.2^n-8.7^n}{4.3^n+5.7^n}\)
b, lim\(\frac{2^{n+1}\left(3.2^n-3^{n-2}\right)}{3^n\left(2^{n-1}+4\right)}\)
c, lim\(\frac{\left(-3\right)^n+2.5^n}{1-5^n}\)
d, lim\(\frac{1+2+3+...+n}{n^2+n+1}\)
Tính
lim \(\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+.....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\right)\)
Tìm các giới hạn sau:
a)\(lim\left[n^2\left(\sqrt{n^2+2}-\sqrt{n^2+4}\right)\right]\)
b)lim( \(\dfrac{3}{n-2}-5n\))
c) lim(\(\dfrac{n-1}{\sqrt{3}-n}-\dfrac{4}{2^{-n}}\))
d) \(lim\left(\dfrac{n^2-4}{n-2}-\dfrac{3n^2+4}{n}\right)\)
e) \(lim\dfrac{\sqrt{n^2+1}-n\sqrt{5}}{\sqrt{n^2+1}+n\sqrt{5}}\)
giới hạn \(lim\dfrac{1-2+4-...+\left(-2\right)^{n-1}}{1-3+9-...+\left(-3\right)^{n-1}}=\dfrac{4\left[1-\left(-2\right)^n\right]}{3\left[1-\left(-3\right)^n\right]}\) bằng?
Cho dãy xác định \(\left\{{}\begin{matrix}u\left(1\right)=\dfrac{1}{4}\\u\left(n+1\right)=\left(u\left(n\right)\right)^2+\dfrac{u\left(n\right)}{2}\end{matrix}\right.\)
CM với mọi n thì 0<u(n)<\(\dfrac{1}{4}\) và\(\dfrac{u\left(n+1\right)}{u\left(n\right)}\le\dfrac{3}{4}\)
Từ đó suy ra limu(n)=o
