cmr:
\(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}....\dfrac{2n-1}{2n}\le\dfrac{1}{\sqrt{3n+1}}\left(\forall n\ge1\right)\)
cmr:
\(\dfrac{1}{9}+\dfrac{1}{16}+...+\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{4}\left(\forall n\ge1\right)\)
Ta có: \(\dfrac{1}{9}=\left(\dfrac{1}{3}\right)^2=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
\(\dfrac{1}{16}=\left(\dfrac{1}{4}\right)^2=\dfrac{1}{4.4}< \dfrac{1}{3.4}\)
................
\(\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{2n\left(2n+1\right)}\)
⇒\(\dfrac{1}{9}+\dfrac{1}{16}+......+\dfrac{1}{\left(2n+1\right)^2}\)< \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.....+\dfrac{1}{2n.\left(2n+1\right)}\)
= \(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{2n}-\dfrac{1}{2n+1}\)
= \(\dfrac{1}{2}-\dfrac{1}{2n+1}\)
= \(\dfrac{2n+1-2}{2n+1}\)
= \(\dfrac{2n-1}{2n+1}\)= \(1-\dfrac{2}{2n+1}\)
Ta có: n ≥ 1⇒ 2n+1 ≥ 3
⇒ \(1-\dfrac{2}{2n+1}\) ≤ \(\dfrac{1}{3}\)
hình như đề sai thì phải
Tính :6/ lim\(\dfrac{-n^2+2n+1}{\sqrt{3n^4+2}}\)
7/ lim \(\dfrac{\sqrt{n^3-2n+5}}{3+5n}\)
10/ lim\(\dfrac{1+3+5+...+\left(2n+1\right)}{3n^3+4}\)
Tìm các giới hạn sau:
a) \(lim\sqrt[3]{-n^3+2n^2-5}\)
b) \(lim\dfrac{1}{\sqrt{n+1}-\sqrt{n}}\)
c) \(lim\left(\dfrac{1}{n+1}-n\right)\)
d) \(lim\left(\dfrac{2n^2-1}{n+1}-2n\right)\)
e) \(lim\dfrac{2n^3+n^2-3n+1}{2-3n}\)
\(a=\lim n\left(\sqrt[3]{-1+\dfrac{2}{n}-\dfrac{5}{n^3}}\right)=+\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n+1}+\sqrt{n}\right)=+\infty\)
\(c=\lim n\left(\dfrac{1}{n^2+n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim\left(\dfrac{2n^2-1-2n\left(n+1\right)}{n+1}\right)=\lim\left(\dfrac{-1-2n}{n+1}\right)=-2\)
\(e=\lim\dfrac{2n^2+n-3+\dfrac{1}{n}}{\dfrac{2}{n}-3}=\dfrac{+\infty}{-3}=-\infty\)
CMR \(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{2n-1}{2n}\le\dfrac{1}{\sqrt{3n+1}}\)
\(\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}......\dfrac{2n-1}{2n}=\dfrac{1.2.3.....\left(2n-1\right)}{2.3.4.....2n}=\dfrac{1}{2n}\)
Khi đó ta có điều cần chứng minh:
\(\dfrac{1}{2n}\le\dfrac{1}{\sqrt{3n+1}}\left(n>\dfrac{1}{3}\right)\)
Hay
\(\dfrac{\sqrt{3n+1}}{2n\left(\sqrt{3n+1}\right)}\le\dfrac{2n}{2n\left(\sqrt{3n+1}\right)}\)
Hay \(\sqrt{3n+1}\le2n\)(luôn đúng)
Tính các giới hạn sau
1,Lim\(\left(\dfrac{2n^3}{2n^2+3}+\dfrac{1-5n^2}{5n+1}\right)\)
2,a,Lim\(\left(\sqrt{n^2+n}-\sqrt{n^2+2}\right)\)
b,Lim\(\dfrac{\sqrt{n^4+3n-2}}{2n^2-n+3}\)
c,Lim\(\dfrac{\sqrt{n^2-4n}-\sqrt{4n^2+1}}{\sqrt{3n^2+1}-n}\)
\(a=\lim\left(\dfrac{2n^3\left(5n+1\right)+\left(2n^2+3\right)\left(1-5n^2\right)}{\left(2n^2+3\right)\left(5n+1\right)}\right)\)
\(=\lim\left(\dfrac{2n^3-13n^2+3}{\left(2n^2+3\right)\left(5n+1\right)}\right)=\lim\dfrac{2-\dfrac{13}{n}+\dfrac{3}{n^3}}{\left(2+\dfrac{3}{n^2}\right)\left(5+\dfrac{1}{n}\right)}=\dfrac{2}{2.5}=\dfrac{1}{5}\)
\(b=\lim\left(\dfrac{n-2}{\sqrt{n^2+n}+\sqrt{n^2+2}}\right)=\lim\dfrac{1-\dfrac{2}{n}}{\sqrt{1+\dfrac{1}{n}}+\sqrt{1+\dfrac{2}{n}}}=\dfrac{1}{2}\)
\(c=\lim\dfrac{\sqrt{1+\dfrac{3}{n^3}-\dfrac{2}{n^4}}}{2-\dfrac{2}{n}+\dfrac{3}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\dfrac{\sqrt{1-\dfrac{4}{n}}-\sqrt{4+\dfrac{1}{n^2}}}{\sqrt{3+\dfrac{1}{n^2}}-1}=\dfrac{1-2}{\sqrt{3}-1}=-\dfrac{1+\sqrt{3}}{2}\)
Tính các tổng sau:
A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+.....+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
B=\(\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}\)
C=\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+....+\sqrt{1+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}\)
\(\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}=\dfrac{1}{3}\left(\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{3n-1}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{3n+2}\right)\)
\(\Rightarrow A=\dfrac{3n}{6\left(3n+2\right)}=\dfrac{n}{6n+4}\)
\(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}=\dfrac{1}{4}\left(\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{3.7}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{1}{4}\left(\dfrac{1}{1.3}-\dfrac{1}{\left(2n+1\right)\left(2n+3\right)}\right)\)
\(\Rightarrow B=\dfrac{n\left(n+2\right)}{3\left(2n+1\right)\left(2n+3\right)}\)
\(\sqrt{1+\dfrac{1}{n^2}+\dfrac{1}{\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n^2+2n+1}{n^2\left(n+1\right)^2}}=\sqrt{\dfrac{n^2\left(n+1\right)^2+2n\left(n+1\right)+1}{n^2\left(n+1\right)^2}}\)
\(=\sqrt{\dfrac{\left[n\left(n+1\right)+1\right]^2}{n^2\left(n+1\right)^2}}=\dfrac{n\left(n+1\right)+1}{n\left(n+1\right)}=1+\dfrac{1}{n\left(n+1\right)}=1+\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow C=1+\dfrac{1}{1}-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+1+\dfrac{1}{3}-\dfrac{1}{4}+...+1+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(\Rightarrow C=2019-\dfrac{1}{2019}\)
CMR \(\dfrac{1.3.5...\left(2n-1\right)}{2.4.6...2n}< \dfrac{1}{\sqrt{2n+1}}\) \(\forall n\in Z_+\)
Lời giải:
Sử dụng quy nạp:
Với \(n=1\Rightarrow \frac{1}{2}< \frac{1}{\sqrt{3}}\) (đúng)
Với \(n=2\Rightarrow \frac{1.3}{2.4}< \frac{1}{\sqrt{5}}\) (đúng)
.............
Giả sử bài toán đúng với \(n=k\), tức là :
\(\frac{1.3.5...(2k-1)}{2.4.6...2k}< \frac{1}{\sqrt{2k+1}}\) (*)
Ta cần chỉ ra nó cũng đúng với \(n=k+1\) hay :
\(\frac{1.3.5....(2k-1)(2k+1)}{2.4.6....(2k)(2k+2)}< \frac{1}{\sqrt{2k+3}}\). Thật vậy, theo (*) ta có:
\(\frac{1.3.5....(2k-1)(2k+1)}{2.4.6....(2k)(2k+2)}< \frac{1}{\sqrt{2k+1}}.\frac{2k+1}{2k+2}=\frac{\sqrt{2k+1}}{2k+2}\) (1)
Xét \(\frac{\sqrt{2k+1}}{2k+2}-\frac{1}{\sqrt{2k+3}}=\frac{\sqrt{(2k+1)(2k+3)}-(2k+2)}{(2k+2)\sqrt{2k+3}}\) \(=\frac{-1}{[\sqrt{(2k+1)(2k+3)}+(2k+2)](2k+2)\sqrt{2k+3}}<0\)
Suy ra \(\frac{\sqrt{2k+1}}{2k+2}< \frac{1}{\sqrt{2k+3}}(2)\)
Từ \((1);(2)\Rightarrow \frac{1.3.5....(2k-1)(2k+1)}{2.4.6....(2k)(2k+2)}< \frac{1}{\sqrt{2k+3}}\)
Vậy bài toán đúng với \(n=k+1\), phép quy nạp hoàn thành.
Do đó ta có đpcm.
Tìm giới hạn các dãy số sau
a) \(lim\dfrac{2^n+6^n-4^{n-1}}{3^n+6^{n+1}}\)
b) \(lim\dfrac{1+3+5+...+\left(2n+1\right)}{3n^2+4}\)
c) \(lim\dfrac{1+2+3+...+n}{n^2-3}\)
d) \(lim\left[\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n+1\right)}\right]\)
e) \(lim\left[\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)}\right]\)
\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)
\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)
\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)
\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)
\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)
\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)
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 \(\left(\sqrt{n^2+3n-1}-\sqrt[3]{n^3-n}\right)\)
a/ Bạn coi lại đề bài, 3n^2 +n^2 thì bằng 4n^2 luôn chứ ko ai cho đề bài như vậy cả
b/ \(\lim\limits\dfrac{\dfrac{n^3}{n^3}+\dfrac{3n}{n^3}+\dfrac{1}{n^3}}{-\dfrac{n^3}{n^3}+\dfrac{2n}{n^3}}=-1\)
c/ \(=\lim\limits\dfrac{-\dfrac{2n^3}{n^2}+\dfrac{3n}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{n}{n^2}}=\lim\limits\dfrac{-2n}{-1}=+\infty\)
d/ \(=\lim\limits\left[n\left(1+1\right)\right]=+\infty\)
e/ \(\lim\limits\left[2^n\left(\dfrac{2n}{2^n}-3+\dfrac{1}{2^n}\right)\right]=\lim\limits\left(-3.2^n\right)=-\infty\)
f/ \(=\lim\limits\dfrac{4n^2-n-4n^2}{\sqrt{4n^2-n}+2n}=\lim\limits\dfrac{-\dfrac{n}{n}}{\sqrt{\dfrac{4n^2}{n^2}-\dfrac{n}{n^2}}+\dfrac{2n}{n}}=-\dfrac{1}{2+2}=-\dfrac{1}{4}\)
g/ \(=\lim\limits\dfrac{n^2+3n-1-n^2}{\sqrt{n^2+3n-1}+n}+\lim\limits\dfrac{n^3-n^3+n}{\sqrt[3]{\left(n^3-n\right)^2}+n.\sqrt[3]{n^3-n}+n^2}\)
\(=\lim\limits\dfrac{\dfrac{3n}{n}-\dfrac{1}{n}}{\sqrt{\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}-\dfrac{1}{n^2}}+\dfrac{n}{n}}+\lim\limits\dfrac{\dfrac{n}{n^2}}{\dfrac{\sqrt[3]{\left(n^3-n\right)^2}}{n^2}+\dfrac{n\sqrt[3]{n^3-n}}{n^2}+\dfrac{n^2}{n^2}}\)
\(=\dfrac{3}{2}+0=\dfrac{3}{2}\)
a) lim \(\left(-3n^3+n^2-1\right)\)
minh le oi ban dao mau so cua ban len cho tu uong roi thay vi tri cua mau thanh n3 +2n