Valid question or not?

Can you check to see if this is even a valid question. I got this challenge from someone and I believe there's no way to solve it but have a look. I can't find a perfect bounded above value...I tried one, no good, tried 2, 3, 4, won't work.



Thanks in advance

This is a calculus type problem.

Given
a1= 3 as in 'a subscript 1'

For n greater than or equal to an+1 = sqrt of 3+an

SHOW an is increasing and bounded above. SHOW an (a subscript n) approaches L and find L
FIND Limit an as n approached infinity

Requires induction for the showing part.

If the attached formula is what you mean, not only that the formula has a bound, it is monotonically decreasing, with a limit at: 2.30277563773199, and highest value at n=1 (a1 = 3).

This can be achieved by assuming there is a finite limit, when approaching the limit, An will be equal to An+1, and therefore:

Code:

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An = sqrt(An + 3) ==>

An^2 -An - 3 = 0 ==>

An = (1 +- sqrt(13)) / 2

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The minus is not an option because the formula is always positive, so:

Code:

---

An = (1 + sqrt(13)) / 2 = 2.30277563773199...

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But I think that this is not the formula you mean...

Regards,
Zachm

How can it be decreasing when the question stated show that its increasing, I sort of understand your way of calculating it but I don't think the formula is right.

You can test this, using MS Excel, for instance, I got the following series:

Code:

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3.00000000000000
2.44948974278318
2.33441421833898
2.30963508337118
2.30426454283599
2.30309889992505
2.30284582634727
2.30279087768457
2.30277894676944
2.30277635622078
2.30277579373694
2.30277567160523
2.30277564508687
2.30277563932895
2.30277563807874
2.30277563780728
2.30277563774834
2.30277563773554
2.30277563773277
2.30277563773216
2.30277563773203
2.30277563773200
2.30277563773200
2.30277563773200
2.30277563773199
2.30277563773199
2.30277563773199
2.30277563773199
2.30277563773199

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I don't know if the question is stated correctly, but if it refers to the formula I attached in my last post, then the series is DECREASING.

Regards,
Zachm