请问∑-2/(2n+1) （n取1到∞）是收敛级数吗
\u5224\u522b\u7ea7\u6570 \u22111\uff01+2\uff01+\u2026n!/\uff082n\uff09\uff01\u7684\u655b\u6563\u6027\u4f60\u597d\uff01\u5f53n>3\u65f6\uff0c1!+2!+\u2026+n!/(2n)!<n(n!)/(2n)!<1/2(2n-2)(2n-1)~(1/8)(1/n^2)\uff0c\u800c\u22111/n^2\u6536\u655b\uff0c\u6839\u636e\u6bd4\u8f83\u5224\u522b\u6cd5\u7684\u6781\u9650\u5f62\u5f0f\u77e5\u22111/2(2n-2)(2n-1)\u6536\u655b\uff0c\u518d\u7531\u6bd4\u8f83\u5224\u522b\u6cd5\u77e5\u22111!+2!+\u2026+n!/(2n)!\u4e5f\u6536\u655b\u3002\u7ecf\u6d4e\u6570\u5b66\u56e2\u961f\u5e2e\u4f60\u89e3\u7b54\uff0c\u8bf7\u53ca\u65f6\u91c7\u7eb3\u3002\u8c22\u8c22\uff01
\u9898\u76ee\u9519\u4e86\u5427\uff0c\u5e94\u662f\u201c 1/(n³+2n²) \u201d\u5427
1/(n³+2n²) < 1/(n³+2n²-3n)
1/(n³+2n²-3n) = 1/[n(n+3)(n-1)]
= (1/2) [(n+3)+(n-1)-2n]/ [n(n+3)(n-1)]
= 1/[2n(n-1)] +1/[2n(n+3)] - 1/[(n+3)(n-1)]
= [1/(n-1) - 1/n]/2 + [1/n - 1/(n+3)]/6 - [1/(n-1) - 1/(n+3)]/4
\u56e0\u6b64\uff0c\u2211(n=1,\u221e) 1/(n³+2n²-3n) \u662f\u6536\u655b\u7684\uff0c
\u5219\u2211(n=1,\u221e) 1/(n³+2n²)\u4e5f\u662f\u6536\u655b\u7684\u3002
用根值判别法得 k=（an）^(1/n)=（2/(2n+1)）^(1/n)=1
则此判别法失效 比值判别法同样失效
又因为an（n趋于∞时）2/(2n+1）趋于n/1 对由1/n构成的p级数 p=1时 近似1/n 所以原级数发散
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