please can you help me I have done my best

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soufiane

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sequence Un=√(n+1) - √n
Vn= 1/√2 + 1/√3 +1/√4 ..........+1/√n
1) prove that 0≤Un≤ 1/2Un

*1) prove that 0≤Un≤ 1/2√n
 
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[MATH]\frac{u_n}{\sqrt{n}}=\sqrt{1+\tfrac{1}{n}}-1\hspace4ex [/MATH](n≥1)
It is clear that this is >0 and decreasing as n increases, so the max is when n=1.
You should be able to get the inequality now.
 
What is Vn supposed to be and how does it relate to the problem?

You say you've done your best... What exactly have you done? If we know that we can help you better.

-Dan
 
soufiane said:
sequence Un=√(n+1) - √n
Vn= 1/√2 + 1/√3 +1/√4 ..........+1/√n

*1) prove that 0 ≤ Un ≤ 1/2√n
Click to expand...


Edit this to:

sequence Un = √(n+1) - √n
Vn= 1/√2 + 1/√3 + 1/√4 + ... + 1/√n

*1) Prove that 0 ≤ Un ≤ 1/(2√n)

\(\displaystyle U_n \ = \ \dfrac{\sqrt{n + 1} - \sqrt{n}}{1}\cdot \dfrac{\sqrt{n + 1} + \sqrt{n}}{\sqrt{n + 1} + \sqrt{n}} \ = \ \dfrac{1}{\sqrt{n + 1} + \sqrt{n}} \)

Can you make a comparison to \(\displaystyle \ \dfrac{1}{2\sqrt{n}} \ \)?


(Edits)
 
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soufiane said:
sequence Un=√(n+1) - √n
Vn= 1/√2 + 1/√3 +1/√4 ..........+1/√n
1) prove that 0≤Un≤ 1/2Un

*1) prove that 0≤Un≤ 1/2√n
Click to expand...
Please show us what you have tried and exactly where you are stuck.

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