Infinite Series: Convergence and Absolute Convergence

fjotlandj
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Given that the series from n = 1 to infinity of An converges to L, which of the following conclusions is valid for the series from n = 1 to infinity of (An)^2?

A) It may diverge
B) It converges absolutely
C) It converges to M < L
D) It converges to M > L
E) It converges to M^2 = L

My intuition tells me the answer is E). But I am not sure can someone help figure this out please
 
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fjotlandj said:
Given that the series from n = 1 to infinity of An converges to L, which of the following conclusions is valid for the series from n = 1 to infinity of (An)^2?

A) It may diverge
B) It converges absolutely
C) It converges to M < L
D) It converges to M > L
E) It converges to M^2 = L

My intuition tells me the answer is E). But I am not sure can someone help figure this out please

Hints: convergent geometric series (this should eliminate some of the possible answers). Then look up what it means to converge absolutely, and play around with

\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\sqrt{n}}
 
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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