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Suppose we have a series $\sum_{n=1}^{\infty}a_n$ such that it is both convergent and absolute convergent. Will the series $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}|a_n|$ converge to the same value ?

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Consider $-1+0+0+\cdots$ and $1+0+0+\cdots$. – David Mitra Sep 20 '13 at 12:30
@DavidMitra this should have been an answer – SztupY Sep 20 '13 at 15:09

No. Take $a_n = \frac{(-1)^{n+1}}{n^2}$. Then $$\sum_{n \ge 1} |a_n| = \frac{\pi^2}{6}$$ (a rough estimate simply shows that the quantity exceeds $1 + \frac{1}{4} = \frac{5}{4}$) while $$\sum_{n \ge 1} a_n < 1$$

As a more general statement, provided that $a_n$ has a negative entry, the quantities will never be equal.

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+1 for the last sentence. – Did Sep 20 '13 at 12:12

Not necessarily. We have: $$ \sum_{n=0}^\infty \left(-\frac{1}{2}\right)^n = \frac{2}{3} $$

However: $$ \sum_{n=0}^\infty \left(\frac{1}{2}\right)^n = 2 $$

Both sums follow from the general formula for geometric series. In general, equality holds only if all terms are non-negative. For suppose $a_{n_0} < 0$, then: $$ \sum_{n=0}^\infty (|a_n| - a_n) \ge |a_{n_0}| - a_{n_0} > 0 $$

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If $a_n$ consists of only positive terms then definitely they will converge to the same limit. For the non-trivial part You will find dozens of Counterexamples : consider these $\sum_{n=0}^\infty \left(-\frac{1}{4}\right)^n and \sum_{n=0}^\infty \left(\frac{1}{4}\right)^n $

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even maybe $\sum_{n=1}^{\infty}a_n$ converges while $\sum_{n=1}^{\infty}|a_n|$ diverges. such as $a_n=\frac{(-1)^{n+1}}{n}$

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