∑k=1∞(−1)k+1(k+3)k(k+1)\sum\limits_{k = 1}^\infty {\frac{{{{( - 1)}^{k + 1}}(k + 3)}}{{k(k + 1)}}}k=1∑∞k(k+1)(−1)k+1(k+3)
1)limk→∞∣ak∣=limk→∞k+3k(k+1)=02)∣ak∣−∣ak+1∣=k+3k(k+1)−k+4(k+1)(k+2)=(k+3)(k+2)−(k+4)kk(k+1)(k+2)==k2+5k+6−k2−4kk(k+1)(k+2)=k+6k(k+1)(k+2)>0⇒∣ak∣>∣ak+1∣\begin{array}{l} 1)\mathop {\lim }\limits_{k \to \infty } |{a_k}| = \mathop {\lim }\limits_{k \to \infty } \frac{{k + 3}}{{k(k + 1)}} = 0\\ 2)|{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)}} = \\ = \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}}| \end{array}1)k→∞lim∣ak∣=k→∞limk(k+1)k+3=02)∣ak∣−∣ak+1∣=k(k+1)k+3−(k+1)(k+2)k+4=k(k+1)(k+2)(k+3)(k+2)−(k+4)k==k(k+1)(k+2)k2+5k+6−k2−4k=k(k+1)(k+2)k+6>0⇒∣ak∣>∣ak+1∣
Then, by Leibniz criterion, this series converges
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