\(1+2+...+n=\frac{n\left(n+1\right)}{2}\Rightarrow\frac{1}{1+2+...+n}=\frac{2}{n\left(n+1\right)}=\frac{2}{n}-\frac{2}{n+1}\)
\(\Rightarrow L=lim\left(1+\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+...+\frac{2}{n}-\frac{2}{n+1}\right)\)
\(=lim\left(2-\frac{2}{n+2}\right)=2-0=2\)
tìm giới hanjn
1) lim \(\frac{\left(-1\right)^n}{n-3}\)
2) lim \(\frac{n\left(sin\left(pi.n^2\right)\right)}{n^2+3n-2}\)
\(\lim\limits\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(\lim\limits\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+n}\right)\)
a; lim\(\frac{\sqrt{6n^4+n+1}}{2n^2+1}\)
b; lim \(\frac{\left(n+1\right)\left(2n+1\right)^2\left(3n+1\right)^3}{n^2\left(n+2\right)^2\left(1-3n\right)^2}\)
lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
lim \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\)
lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
\(lim\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+......+\frac{1}{\sqrt{n^2+n}}\right)\)
Cho dãy số \(\left(u_n\right)\) : \(\left\{{}\begin{matrix}u_1=\frac{5}{2}\\u_{n+1}=\frac{1}{2}u_n^2-u_n+2\end{matrix}\right.\) với n=1,2,3... Chứng minh rằng \(\lim\limits_{n\rightarrow+\infty}u_n=+\infty\) và tìm \(\lim\limits_{n\rightarrow+\infty}\left(\frac{1}{u_1}+\frac{1}{u_2}+...+\frac{1}{u_n}\right)\) ?
lim (\(\frac{1}{2}\)+ \(\frac{\left(-1\right)^n.\left(\sqrt{3}\right)^n}{3.2^{n+1}}\))
a)lim \(\frac{\left(2n+1\right)^2\left(n-1\right)}{\sqrt[3]{n^3+7n-2}}\)
b)lim [(2n-1)\(\sqrt{\frac{2n^2+5}{n^4+n^2+2}}\)]
c)lim [n(\(\sqrt[3]{n^3+n^2}-n\))]
