Proof of the Ratio Test and the Triangle Inequality

Fiz2007
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Homework Statement



Prove: If the limit inf as k goes to infinity of abs(ak+1 / ak) > 1 then the sum from 1 to infinity of ak diverges


Homework Equations





The Attempt at a Solution



So far I have this:

Suppose lim inf abs(ak+1/ak) >1
then, there exists an r such that lim inf abs(ak+1/ak) > r > 1
then \exists N an element of the natural numbers such that k >= N implies
abs(ak+1/ak) > r
that is, for k >= N, abs(ak+1) > abs(ak) r
and, abs (an+1) > r abs(an)
abs (an+2) > r abs (an+1) > r^2 abs(an)
and in general, abs (an+k) > r^k abs(an)

the series the sum from 1 to infinity of abs(an) * r^k diverges to infinity (geometric series with r>1).

Therefore the sum from 1 to infinity of abs(an+k) diverges to infinity by the comparison test.

And then I'm stuck... I'm not sure how to go from the absolute value diverging to the series diverging. It clearly does not converge absolutely but what about conditional convergence? Please help!

Thanks
 
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A necessary condition for convergence of a series is a_n approaches to 0 at infinity.
 
Thanks! that made it much simpler.
 
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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