Limit of a Sequence with (-1)n and Convergence Analysis

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Homework Statement



... I have (-1)n in a sequence and I'm trying to find the limit of that sequence?

Homework Equations



Obviously we have -1, 1, -1, 1, ...

The Attempt at a Solution



You see, I'm trying to consider the case xn = n[1+(-1)n] + (1/n), to consider whether the sequence has just one accumulation point and is convergent.
 
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The sequence does seem to have one accumulation point, but could it converge? Write out the first few terms.
 
Tedjn said:
The sequence does seem to have one accumulation point, but could it converge? Write out the first few terms.

xn = {1, 9/4, 1/3, 65/8, 1/5, 125/12, 1/7, ...}
 
If a sequence is of the form (-1)^n a_n where a_n\ge 0 for all n, then either the sequence does not converge or it converges to 0.

Proof: suppose it converged to a> 0. Let \epsilon= a/2. Then for any N> 0, there exist odd n> N so that (-1)^na_n< 0 from which |a- (-1)^n a_n|> a and not less than \epsilon= a/2.

You do the case for a< 0.
 
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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