"\\sum\\limits_{k = 1}^\\infty {\\frac{{{{( - 1)}^{k + 1}}(k + 3)}}{{k(k + 1)}}}"
"\\begin{array}{l}\n1)\\mathop {\\lim }\\limits_{k \\to \\infty } |{a_k}| = \\mathop {\\lim }\\limits_{k \\to \\infty } \\frac{{k + 3}}{{k(k + 1)}} = 0\\\\\n2)|{a_k}| - |{a_{k + 1}}| = \\frac{{k + 3}}{{k(k + 1)}} - \\frac{{k + 4}}{{(k + 1)(k + 2)}} = \\frac{{(k + 3)(k + 2) - (k + 4)k}}{{k(k + 1)(k + 2)}} = \\\\\n = \\frac{{{k^2} + 5k + 6 - {k^2} - 4k}}{{k(k + 1)(k + 2)}} = \\frac{{k + 6}}{{k(k + 1)(k + 2)}} > 0 \\Rightarrow |{a_k}| > |{a_{k + 1}}|\n\\end{array}"
Then, by Leibniz criterion, this series converges
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