Question

[Vương Nguyên]

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

[Akai Haruma]

$n$ tiến đến đâu vậy bạn?

Answer 2

[Akai Haruma]

Câu 2:

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

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

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

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