Question

Đã trả lời bởi chuyên gia

Answer 1

Với mọi n ∈ ℕ* ta có

\frac{1}{(n + 1) \sqrt{n}} < 2 (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})

\Leftrightarrow 1 < 2 (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}}) (n + 1) \sqrt{n}

\Leftrightarrow 1 < 2 (n + 1) - 2 \sqrt{n (n + 1)}

\Leftrightarrow n + 1 - 2 \sqrt{n (n + 1)} + n > 0

$\Leftrightarrow {(\sqrt{n + 1} - \sqrt{n})}^{2} > 0$ (luôn đúng)

Áp dụng bất đẳng thức trên ta có

\frac{1}{2 \sqrt{1}} < 2 (\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}})

\frac{1}{3 \sqrt{2}} < 2 (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}})

...

\frac{1}{2005 \sqrt{2004}} < 2 (\frac{1}{\sqrt{2004}} - \frac{1}{\sqrt{2005}})

Suy ra

$\frac{1}{2 \sqrt{1}} + \frac{1}{3 \sqrt{2}} + \frac{1}{4 \sqrt{3}} + ... + \frac{1}{2005 \sqrt{2004}} < 2 (\frac{1}{\sqrt{1}} - \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{2004}} - \frac{1}{\sqrt{2005}})$

$\Leftrightarrow \frac{1}{2 \sqrt{1}} + \frac{1}{3 \sqrt{2}} + \frac{1}{4 \sqrt{3}} + ... + \frac{1}{2005 \sqrt{2004}} < 2 (1 - \frac{1}{\sqrt{2005}}) < 2$

Vậy $\frac{1}{2 \sqrt{1}} + \frac{1}{3 \sqrt{2}} + \frac{1}{4 \sqrt{3}} + ... + \frac{1}{2005 \sqrt{2004}} < 2$ .

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Answer 2