How to finish ths last step of the induction question

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here is the full question and the last step to which i understand
i don't know how to go further??

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i got this suggestion from hallsofivy :
You know that for every n, a_n- L- \epsilon< a_{n+1}< an+ L+ \epsilon so a_n- k(L+ \epsilon)< a_{n+k}< a_n+ k(L+ \epsilon) is certainly true for k= 1. Now suppose a_n- k(L+ \epsilon)< a_n< a_{n+1}+ k(L+ \epsilon) is true for some specific k and all n. Then a_n- (k+1)(L+ \epsilon)= [a_n- (L+ \epsilon)]- k(L+\epsilon)< a_{n+1}-k(L+\epsilon) and now use a_n- k(L+ \epsilon)< a_{n+1} with n+1 instead of n- which you can do because it is true for all n.


but i can't see in it my base (k) expression and using it to prove the (k+1) expression
i don't know how to apply it the step i got stuck
??
 
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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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