Question

Tính:

Câu 1: lim ( \(\frac{1}{\sqrt{n^2+1}}\) + \(\frac{1}{\sqrt{n^2+2}}\) + ... + \(\frac{1}{\sqrt{n^2+n}}\) )

Câu 2: lim ( \(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) +...+ \(\frac{1}{n\left(n+1\right)}\) )

Câu 3: lim ( \(\frac{1}{n^2}\) + \(\frac{3}{n^2}\) + \(\frac{5}{n^2}\) +...+ \(\frac{2n-1}{n^2}\) )

Câu 4: lim ( \(\sqrt{3+\frac{n^2-1}{3+n^2}}\) - \(\frac{\left(-1\right)^n}{2^n}\) )

Câu 5: lim \(\sqrt{\frac{cos2n}{3n}+9}\)

Answer 1

$n$ tiến đến đâu vậy bạn?

Answer 2

Câu 2:

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{(n+1)-n}{n(n+1)}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...\frac{1}{n}-\frac{1}{n+1}\)

\(=1-\frac{1}{n+1}\)

\(\Rightarrow \lim_{n\to \infty}(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)})=\lim_{n\to \infty}(1-\frac{1}{n+1})=1-\lim_{n\to \infty}\frac{1}{n+1}=1-0=1\)

Answer 3

Câu 3:

Ta biết rằng $\lim_{x\to \infty}\frac{1}{x}=0\Rightarrow \lim_{x\to \infty}\frac{a}{x}=0$ với $a\in\mathbb{R}$

Do đó:

$\lim_{n\to \infty}\frac{1}{n^2}=0$

$\lim_{n\to \infty}\frac{2}{n^2}=0$

.....

$\lim_{n\to \infty}\frac{2n-1}{n^2}=\lim_{n\to \infty}(\frac{2}{n}-\frac{1}{n^2})=\lim_{n\to \infty}\frac{2}{n}-\lim_{n\to \infty}\frac{1}{n^2}=0-0=0$

Do đó:

$\lim_{n\to \infty}(\frac{1}{n^2}+...+\frac{2n-1}{n^2})=\lim_{n\to \infty}\frac{1}{n^2}+....+\lim_{n\to \infty}\frac{2n-1}{n^2}=0+0+...+0=0$

Answer 4

Câu 4:

\(\lim_{n\to \infty}\sqrt{3+\frac{n^2-1}{3+n^2}}=\lim_{n\to \infty}\sqrt{3+1-\frac{4}{n^2+3}}=\sqrt{4-0}=2\)

\(\lim_{n\to \infty}\frac{(-1)^n}{2^n}=\lim_{n\to \infty}(\frac{-1}{2})^n=0\) (định lý đã có trong SGK)

\(\Rightarrow \lim_{n\to \infty}(\sqrt{3+\frac{n^2-1}{n^2+3}}-\frac{(-1)^n}{2^n})=2-0=2\)

Answer 5

Câu 5:

Ta thấy $\cos 2n$ là hàm bị chặn với mọi $n\to \infty$

$\lim_{n\to \infty}\frac{1}{3n}=0$

$\Rightarrow \lim_{n\to \infty} \sqrt{\frac{\cos 2n}{3n}+9}=\sqrt{0+9}=3$

Answer 6

Câu 1:

ĐK: $n$ nguyên dương
Ta thấy:

$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}< \frac{1}{\sqrt{n^2}}+\frac{1}{\sqrt{n^2}}+...+\frac{1}{\sqrt{n^2}}=\frac{n}{\sqrt{n^2}}=1$

Và:

$\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}> \frac{1}{\sqrt{n^2+n}}+...+\frac{1}{\sqrt{n^2+n}}=\frac{n}{\sqrt{n^2+n}}$

Trong đó $\lim_{n\to +\infty}\frac{n}{\sqrt{n^2+n}}=\lim_{n\to +\infty}\frac{1}{\sqrt{1+\frac{1}{n}}=1$

Do đó theo định lý kẹp ta suy ra:

\(\lim_{n\to +\infty}(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}})=1\)