# Ngô Quốc Anh

## October 18, 2007

### Một số dạng bài tập xét sự hội tụ của chuỗi số dương

Filed under: Giải Tích 4 — Ngô Quốc Anh @ 14:54

1. Let $\sum a_n$ be a convergent series, with $a_n \geq 0$. What can we say about the nature of each of the following series?

 (a) $\displaystyle\sum\sqrt {a_n}$ (b) $\displaystyle\sum a_n^2$ (c) $\displaystyle\sum \frac {a_n}{1 + a_n}$ (d) $\displaystyle\sum \frac {a_n^2}{1 + a_n^2}$ (e) $\displaystyle\sum \frac {\sqrt {a_n}}{n}$ (f) $\displaystyle\sum \frac {a_n}{\sqrt {n}}$ (g) $\displaystyle\sum \frac {a_n}{n}$ (h) $\displaystyle\sum na_n$ (i) $\displaystyle\sum \cos(a_n)$ (j) $\displaystyle\sum \sin(a_n)$ (k) $\displaystyle\sum \tan(a_n)$ (l) $\displaystyle\sum \cos(\sqrt {a_n})$ (m) $\displaystyle\sum [1 - \cos(\sqrt {a_n})]$ (n) $\displaystyle\sum [1 - \cos(a_n)]$ (o) $\displaystyle\sum \arcsin(a_n)$ (p) $\displaystyle\sum \arccos(a_n)$ (q) $\displaystyle\sum a_n^{3/2}$ (r) $\displaystyle\sum a_n^3$ (s) $\displaystyle\sum \frac {a_1 + a_2 + ... + a_n}{n}$ (t) $\displaystyle\sum \arctan(a_n)$ (u) $\displaystyle\sum \frac {1}{1 + a_n}$ (v) $\displaystyle\sum ( - 1)^na_n$ $\displaystyle\sum \sqrt [n]{a_n}$ (x) $\displaystyle\sum (a_n - a_{n + 1})$ (y) $\displaystyle\sum [\ln(1 + a_n)]^{5/4}$

2) Regarding the same series of the previous question, how about if $\sum a_n$ diverges?

Hint.

$\displaystyle\left( a \right) \to a_n = \frac{1}{{n^2 }}$.

$\displaystyle\left( b \right) \to a_n^2 \leqslant a_n \sum\limits_{n = 0}^{ + \infty } {a_n }$.

$\displaystyle\begin{gathered}\left( d \right) \to \frac{{a_n^2 }}{{1 + a_n^2 }} \leqslant \frac{{a_n^2 }}{{2a_n }} = a_n \hfill \\\left( c \right) \to \frac{{a_n }}{{1 + a_n }} \leqslant a_n . \hfill \\\end{gathered}$

$\displaystyle\left( e \right) \to \frac{{\sqrt {a_n } }}{n} \leqslant \frac{1}{2}\left( {a_n + \frac{1}{{n^2 }}} \right)$

$\displaystyle\left( f \right) \to \mathop {\lim }\limits_{n \to \infty } \frac {{\frac {{a_n }} {{\sqrt n }}}} {{a_n }} = \mathop {\lim }\limits_{n \to \infty } \frac {1} {{\sqrt n }} = 0= \mathop {\lim }\limits_{n \to \infty } \frac {1} {n} = \mathop {\lim }\limits_{n \to \infty } \frac {{\frac {{a_n }} {n}}} {{a_n }} \leftarrow \left( g \right)$

$\displaystyle\left( h \right) \to a_n = \frac{1}{{n^2 }}$.

$\displaystyle\left( i \right) \to$.

$\displaystyle\left( j \right) \to |\sin(a_n)| \leqslant a_n$.

$\displaystyle\left( k \right) \to$.

$\displaystyle\left( l \right) \to$.

$\displaystyle\left( m \right) \to$.

$\displaystyle\left( n \right) \to$.

$\displaystyle\left( o \right) \to$.

$\displaystyle\left( p \right) \to$.

$\displaystyle\begin{gathered} \left( q \right) \to a_n^{\frac{3}{2}} \leqslant a_n \hfill \\\left( r \right) \to a_n^3 \leqslant a_n \hfill \\\end{gathered}$

$\displaystyle\left( s \right) \to \frac{{a_1 }}{n} \leqslant \frac{{a_1 + ... + a_n }}{n}$.

$\displaystyle\left( t \right) \to$.

$\displaystyle\left( u \right) \to a_n = \frac{1}{{n^2 }}$.

$\displaystyle\left( v \right) \to$.

$\displaystyle\left( w \right) \to a_n = \frac{1}{{n^n }}$.

$\displaystyle\left( x \right) \to |a_n-a_{n+1}| \leqslant a_n + a_{n+1}$.

$\displaystyle\left( y \right) \to$.

## 6 Comments »

1. giai di. sao co moi bai tap ma khong co loi giai. thi ai biet dau ma lam ha

Comment by giai — May 16, 2009 @ 15:54

• Có gợi ý rồi mà 🙂

Comment by Ngô Quốc Anh — December 16, 2009 @ 22:48

2. giải thích rõ hơn đi bạn…

Comment by tung hee — December 4, 2011 @ 21:04

• Chào tung hee, đó là các dạng, đề bài phát biểu quá rõ ràng rồi, còn cần giải thích gì nữa.

Comment by Ngô Quốc Anh — December 4, 2011 @ 21:14

3. giai thich dum minh cau a @@

Comment by quang — February 16, 2012 @ 22:28

• Câu a) ta không thể khẳng định được chuỗi chưa căn hội tụ. Phản ví dụ như đề cập ở dưới, dùng $\frac{1}{n^2}$.

Comment by Ngô Quốc Anh — February 16, 2012 @ 23:57

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