Increasing or Decreasing Sequence

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A sequence (an) is defined recursively by a1 =1 and an+1 = 1/ 2+an for all n is greater than 1 or equal 1.

ı'll prove that this sequence is convergent,buy ı cannot decide whethet it is increasing or decreasing .When ı write terms ,some terms increase some terms decrease.
 
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What are the first couple of terms that you got? From the definition in your post it seems like the sequence is increasing to me
 
a1=1/3,a2=7/21,a3=3/7,a4=7/17,a5=17/41 ...

according to my calculation, a1>a2, a3>a2, a5>a4, a3>a4

How can ı decide?
 
I suspect that your OP is not very well formatted now. Is the sequence
an+1 = (1/2)+an (which is what you wrote using proper order of operations)

or is it

an+1=1/(2+an)
which seems to match your most recent post

If it's the latter, then just like you described in your post it's neither increasing nor decreasing
 
My sequence is the second one.If it is neither increasing or decreasing ,how will ı show that it is convergent sequence? ı think there is a problem about sequence.

Thank you for efforts :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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