Show that a Sequence is monotonically decreasing

AI Thread Summary
To show that the sequence an = 1/√n is monotonically decreasing, it is necessary to demonstrate that an ≥ an+1. This can be achieved by comparing the magnitudes of √n and √(n+1), noting that √(n+1) is greater than √n. Consequently, the reciprocals indicate that 1/√(n+1) is less than 1/√n, confirming that an ≥ an+1. The discussion emphasizes the importance of maintaining the inequality while manipulating the terms. Overall, the sequence is confirmed to be monotonically decreasing.
Calu
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Homework Statement



I was wondering how I would go about showing that (an) is monotone decreasing given that an = 1/√n.

I believe I have to show an ≥ an+1, but I'm not sure how to go about doing that.
 
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Calu said:

Homework Statement



I was wondering how I would go about showing that (an) is monotone decreasing given that an = 1/√n.

I believe I have to show an ≥ an+1, but I'm not sure how to go about doing that.

Well, PF rules require you to make a start on your own. What do you know about the magnitudes of ##\sqrt{n}## and ##\sqrt{n+1}##?
 
Ray Vickson said:
Well, PF rules require you to make a start on your own. What do you know about the magnitudes of ##\sqrt{n}## and ##\sqrt{n+1}##?
I know that the magnitude of ##\sqrt{n+1}## is larger than that of ##\sqrt{n}##. Therefore I would assume that the opposite would be true for the magnitude of their reciprocals which would make an≥an+1 as required, however I'm not sure how to write this in a more coherent way.
 
Calu said:
I know that the magnitude of ##\sqrt{n+1}## is larger than that of ##\sqrt{n}##. Therefore I would assume that the opposite would be true for the magnitude of their reciprocals which would make an≥an+1 as required, however I'm not sure how to write this in a more coherent way.

(1) \sqrt{n + 1} > \sqrt{n}.
(2) If n > 0 then dividing both sides by \sqrt{n} > 0 preserves the inequality. Hence \frac{\sqrt{n + 1} }{\sqrt{n}} > 1.
(3) Dividing both sides by ... > 0 preserves the inequality. Hence ...
 
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