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Dãy số - Giới hạn

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Topic	Stats	Last Post Info
$\begin{cases} u_1 = 1 \\ u_{n+1} = \sqrt{u_n(u_n+1)(u_n+2)(u_n+3) + 1} \end{cases}$ Started by Ruka, 18-03-2023	1 reply 433 Views	$$\begin{cases} u_1 = 1 \\ u_{n+1} = \sqrt{u_n(u_n+1)(u_n+2)(u_n+3) + 1} \end{cases}$ - last post by William Nguyen$ William Nguyen 12-06-2023
$A_1=2 \in R , A_{n+1}=A_n+\frac{1-A_n^2}{n}, \forall n \ge 1$ Started by do viet anh, 07-06-2023 dãy số	0 replies 340 Views	$$A_1=2 \in R , A_{n+1}=A_n+\frac{1-A_n^2}{n}, \forall n \ge 1$ - last post by do viet anh$ do viet anh 07-06-2023
Tìm miền hội tụ của $\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{n}} $ Started by NguyenHieu11, 29-05-2023	0 replies 240 Views	$Tìm miền hội tụ của $\sum\limits_{n = 1}^\infty {\frac{{{x^n}}}{n}} $ - last post by NguyenHieu11$ NguyenHieu11 29-05-2023
Cho hàm số $f(x)=\frac{1}{2+e^{\frac{1}{x}}}$. Tính $\lim_{x\rightarrow 0}f(x)$ Started by acbd, 09-05-2023 gioihan	1 reply 286 Views	$Cho hàm số $f(x)=\frac{1}{2+e^{\frac{1}{x}}}$. Tính $\lim_{x\rightarrow 0}f(x)$ - last post by minh2906$ minh2906 21-05-2023
$\lim_{x \to 0}\dfrac{1-\sqrt{1+x^4}\cos x^2}{x^3\arctan x^5}$ Started by Thegooobs, 18-05-2023	0 replies 265 Views	$$\lim_{x \to 0}\dfrac{1-\sqrt{1+x^4}\cos x^2}{x^3\arctan x^5}$ - last post by Thegooobs$ Thegooobs 18-05-2023
$u(x) \sim v(x), x \to a, u(x) \ne 0 \Rightarrow v(x) \sim u(x), x \to a $ Started by Thegooobs, 02-05-2023	0 replies 211 Views	$$u(x) \sim v(x), x \to a, u(x) \ne 0 \Rightarrow v(x) \sim u(x), x \to a $ - last post by Thegooobs$ Thegooobs 02-05-2023
$$\lim_{x \to +\infty}\dfrac{2.4^x+3^x+2^x+1}{2.4^x-3^x-2^x-1}$$ Started by Thegooobs, 29-04-2023	0 replies 285 Views	$$$\lim_{x \to +\infty}\dfrac{2.4^x+3^x+2^x+1}{2.4^x-3^x-2^x-1}$$ - last post by Thegooobs$ Thegooobs 29-04-2023
$\lim_{x \to 0} \dfrac{e^x-\cos{x}+ax+bx^2 }{x^3}=c$ Started by Thegooobs, 26-04-2023 giới hạn	1 reply 448 Views	$$\lim_{x \to 0} \dfrac{e^x-\cos{x}+ax+bx^2 }{x^3}=c$ - last post by chuyenndu$ chuyenndu 26-04-2023
$x_{1} = 3, x_{n+1} = {x_{n}}^{2} - 2, n \geq 1 $. Tính $lim \frac{\prod x_{k}}{x_{k+1}}$ Started by tuanduc, 22-04-2023	1 reply 226 Views	$$x_{1} = 3, x_{n+1} = {x_{n}}^{2} - 2, n \geq 1 $. Tính $lim \frac{\prod x_{k}}{x_{k+1}}$ - last post by chuyenndu$ chuyenndu 25-04-2023
$L=\lim_{n \to +\infty }\sum_{k=1}^n 3^{k-1}\sin^3\frac{\alpha}{3^k}$ Started by thinhisthenumber1, 22-04-2023	2 replies 323 Views	$$L=\lim_{n \to +\infty }\sum_{k=1}^n 3^{k-1}\sin^3\frac{\alpha}{3^k}$ - last post by Thegooobs$ Thegooobs 22-04-2023
Số hạng tổng quát dãy không chính phương… Started by hxthanh, 08-04-2023	5 replies 465 Views	hxthanh 09-04-2023
Answered $\begin{cases} u_1 = 2017\\u_{n+1} = \dfrac{2015n+1}{2016n+1}(u_n + 1) \end{cases}$ Started by Ruka, 20-03-2023	3 replies 531 Views	$$\begin{cases} u_1 = 2017\\u_{n+1} = \dfrac{2015n+1}{2016n+1}(u_n + 1) \end{cases}$ - last post by hung2401$ hung2401 21-03-2023
Answered $\begin{cases} x_1 = 3\\x_{n+1} = \dfrac{1}{2}(\sqrt{x_n^2 + 5x_n} + x_n \end{cases}$ Started by Ruka, 10-03-2023	1 reply 1447 Views	$$\begin{cases} x_1 = 3\\x_{n+1} = \dfrac{1}{2}(\sqrt{x_n^2 + 5x_n} + x_n \end{cases}$ - last post by Ruka$ Ruka 10-03-2023
$\begin{cases} u_0=0\\u_{n+1} = \dfrac{u_n + 2008}{-u_n+2010} \end{cases}$ Started by Ruka, 14-02-2023 dãy số	1 reply 396 Views	$$\begin{cases} u_0=0\\u_{n+1} = \dfrac{u_n + 2008}{-u_n+2010} \end{cases}$ - last post by Ruka$ Ruka 07-03-2023
Cho dãy số $(x_n)$ : $\begin{cases} x_1 = 3\\ x_{n+1} = \dfrac{x_n^3 + 2x_n^2 + 4}{x_n^2 - x_n + 6} \end{cases}$ Started by Ruka, 05-03-2023	1 reply 312 Views	$Cho dãy số $(x_n)$ : $\begin{cases} x_1 = 3\\ x_{n+1} = \dfrac{x_n^3 + 2x_n^2 + 4}{x_n^2 - x_n + 6} \end{cases}$ - last post by Ruka$ Ruka 07-03-2023
CTTQ của $u_1=1; u_{n+1}=u_{n}+2n^2$ Started by HongTham2211, 06-12-2017	1 reply 493 Views	$CTTQ của $u_1=1; u_{n+1}=u_{n}+2n^2$ - last post by Ruka$ Ruka 06-03-2023
${S_n}=\sum\limits_{k=1}^n{\frac1{a_k^2+a_k+1}}$ Started by hungnolan, 15-03-2018 ffsdf	2 replies 601 Views	$${S_n}=\sum\limits_{k=1}^n{\frac1{a_k^2+a_k+1}}$ - last post by Ruka$ Ruka 06-03-2023
$\frac{U_{n+2}}{n+2}=\frac{U_{n+1}}{n+1}+\frac{2U_n}{n}$ Started by Rias Gremory, 11-03-2018	2 replies 479 Views	$$\frac{U_{n+2}}{n+2}=\frac{U_{n+1}}{n+1}+\frac{2U_n}{n}$ - last post by Ruka$ Ruka 06-03-2023
$u_{n+1}=(1+\frac{3}{n})u_{n}+2-\frac{3}{n}$ Started by coconut00, 10-08-2021	1 reply 519 Views	$$u_{n+1}=(1+\frac{3}{n})u_{n}+2-\frac{3}{n}$ - last post by Ruka$ Ruka 06-03-2023
Tìm giới hạn dãy số cho bởi công thức truy hồi $u_{n}(3+u_{n-1})+1=0$ Started by ThichHocToancom, 28-11-2021 dãy số, giới hạn and 1 more...	2 replies 580 Views	$Tìm giới hạn dãy số cho bởi công thức truy hồi $u_{n}(3+u_{n-1})+1=0$ - last post by Ruka$ Ruka 06-03-2023
tìm cttq của $u_{n}$, biết $\left\{\begin{matrix} u_{1}=1 & \\ u_{n+1}=\sqrt{\frac{2}{3}u_{n}^2+\frac{n-2}{n^2+n}} & \end{matrix}\right.$ Started by lucas123, 15-08-2021	2 replies 544 Views	$tìm cttq của $u_{n}$, biết $\left\{\begin{matrix} u_{1}=1 & \\ u_{n+1}=\sqrt{\frac{2}{3}u_{n}^2+\frac{n-2}{n^2+n}} & \end{matrix}\right.$ - last post by Ruka$ Ruka 06-03-2023
Một giới hạn suy ra từ bài toán Basel Started by thangthien, 22-02-2023	1 reply 279 Views	thangthien 23-02-2023
$\begin{cases} u_1=2\\u_{n+1} -u_n= \dfrac{1}{2023}(u_n^2-u_n) \end{cases}$ Started by Ruka, 14-02-2023 dãy số	5 replies 546 Views	$$\begin{cases} u_1=2\\u_{n+1} -u_n= \dfrac{1}{2023}(u_n^2-u_n) \end{cases}$ - last post by Ruka$ Ruka 17-02-2023
Dãy số tăng và chặn trên bởi $a$ . $lim u_n = a$ là đúng hay sai? Started by Ruka, 18-01-2023	3 replies 344 Views	perfectstrong 20-01-2023
$x_{1} = \frac{2020}{2019}; x_{n+1} = 2019x_{n}^{2} + x_{n}.$ Started by pmt22042003, 07-03-2019 giới hạn dãy số	3 replies 1291 Views	$$x_{1} = \frac{2020}{2019}; x_{n+1} = 2019x_{n}^{2} + x_{n}.$ - last post by chanhquocnghiem$ chanhquocnghiem 18-01-2023
Tính $\lim_{x\rightarrow +\infty }\frac{1}{2n+1}\sum_{i=1}^{n}\frac{u_{i}^{2}}{1+u_{i}^{2}}$ Started by Le Tuan Canhh, 18-10-2022	4 replies 1196 Views	$Tính $\lim_{x\rightarrow +\infty }\frac{1}{2n+1}\sum_{i=1}^{n}\frac{u_{i}^{2}}{1+u_{i}^{2}}$ - last post by Ruka$ Ruka 15-01-2023
$U_{n+2}=(n+1)(U_{n+1}+U_{n})$ Started by hoangkimca2k2, 23-02-2018 công thức tổng quát	3 replies 696 Views	$$U_{n+2}=(n+1)(U_{n+1}+U_{n})$ - last post by Ruka$ Ruka 15-01-2023
tìm 4 số hạng đầu của 1 cấp số nhân... Started by phamtranbaotram, 16-06-2018	1 reply 290 Views	Ruka 15-01-2023
giả sử các số 5x-y, 2x+3y và x+2y lập thành 1 cấp số cộng..... Started by phamtranbaotram, 16-06-2018	1 reply 559 Views	Ruka 15-01-2023
$frac{U_{n+1}}=\frac{5U_{n}+4}{U_{n}+2}$ Started by buithihatien, 11-03-2018	1 reply 371 Views	$$frac{U_{n+1}}=\frac{5U_{n}+4}{U_{n}+2}$ - last post by Ruka$ Ruka 14-01-2023

Dãy số - Giới hạn

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