Alternating series test for convergence

Maddie1609
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Homework Statement



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Homework Equations



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The Attempt at a Solution



I don't get how they got what's stated in the above picture. Where does 1/2 and n/(n + 1) come from? Can't you just show that an + 1 ≤ an?
 
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Maddie1609 said:
Where does 1/2 and n/(n + 1) come from? Can't you just show that an + 1 ≤ an?
That's what they do. The logic is easier to follow if you start with the result and reduce this to something true (look at the steps in reverse order), but this direction works as well. The first line is then a clever guess what will be needed later (and the statement is clearly true).
 
mfb said:
That's what they do. The logic is easier to follow if you start with the result and reduce this to something true (look at the steps in reverse order), but this direction works as well. The first line is then a clever guess what will be needed later (and the statement is clearly true).
A lot easier to follow that direction, thanks! The final step has 2n and 2n - 1 instead of (-2)n and (-2)n - 1 which I don't get.
 
The test compares the magnitude of the sequence elements only, powers of -1 don't change the magnitude.
The sign has to be checked separately (easy here).
 
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mfb said:
The test compares the magnitude of the sequence elements only, powers of -1 don't change the magnitude.
The sign has to be checked separately (easy here).
Oh okay! Thank you :smile:
 
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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