Bài 102446

Question

Chứng minh với mọi số nguyên dương n:
a) \( \displaystyle
\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\leq 2\sqrt{n}-1
\)
b) \( \displaystyle
\frac{1}{2}\times \frac{3}{4}\times \ldots \times\frac{2n-1}{2n}\leq \frac{1}{\sqrt{2n}}
\)

score 0 · Answer 1

a) Ta có: \(
\frac{1}{\sqrt{k}}=\frac{2}{\sqrt{k}+\sqrt{k}}\leq \frac{2}{\sqrt{k}+\sqrt{k-1}}=2\left ( \sqrt{k}-\sqrt{k-1} \right )
\) Do đó:
\(
\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}\leq \frac{1}{\sqrt{1}}+2\left ( \sqrt{2}-\sqrt{1} \right )+...+2\left ( \sqrt{n}-\sqrt{n-1} \right )\)
\(=1-2+2\sqrt{n}=2\sqrt{n}-1 \).
b) \(
\frac{2k-1}{2k}=\sqrt{\frac{\left (2k-1 \right )^{2}}{\left ( 2k\right )^{2}}}\leq\sqrt{\frac{\left (2k-1 \right )^{2}}{\left ( 2k\right )^{2}-1}} =\sqrt{\frac{2k-1}{2k+1}}
\) do đó:
\(
\frac{1}{2}\times \frac{3}{4}\times ...\times \frac{2n-1}{2n}\leq \sqrt{\frac{1}{3}}\times \sqrt{\frac{3}{5}}\times ...\times \sqrt{\frac{2n-1}{2n+1}}=\frac{1}{\sqrt{2n+1}}\leq \frac{1}{\sqrt{2n}}
\)

Bài 102446

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Bài 102446

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