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Since the series is alternating, it is a candidate for the Leibniz test. Now ${a}_{n}\=1\/\left(2nplus;5\right)$, which is monotone decreasing to zero as $n\to \infty$. Hence, by this test, the series converges conditionally.${}$
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However, the Limit-Comparison test, with the harmonic series as $\mathrm{\Σ}{b}_{n}$, gives
$c\=\underset{n\to \infty}{lim}\frac{{a}_{n}}{{b}_{n}}$ = $\underset{n\to \infty}{lim}\frac{1\/\left(2nplus;5\right)}{1sol;n}$ = $\underset{n\to \infty}{lim}\frac{n}{2nplus;5}equals;1sol;2gt;0$
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Since $c\=1\/2\>0$, and $\mathrm{\Σ}{b}_{n}$ (the harmonic series) diverges, it follows that $\mathrm{\Σ}{a}_{n}$ also diverges. Hence, the given alternating series does not converge absolutely. It only converges conditionally.
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