Would the converse of this be true?

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Homework Statement



Theorem X. If the series ∑un is absolutely convergent, then each of the seires ∑an, ∑bn is convergent, and ∑un = ∑an - ∑bn. [...]

Homework Equations



Meh

The Attempt at a Solution



I was wondering if the converse is true: If I have a series ∑un, and if I can express it as the difference between two convergent sequences, then is ∑un convergent?

For example, one of the homework problems asks whether

1 - 1/32 + 1/52 - 1/72 + ...

is absolutely convergent.

Well, if I take the absolute value of those terms, I have

1 + 1/32 + 1/52 + 1/72 + ...

and I know ∑1/n2 is convergent (from a previous section of the book), so of course any subsequence is convergent.
 
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Yes, the converse is indeed true.
And you also solved your homework problem correctly!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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