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_1714110183822_0第二版第十一章级数训练自测
题
快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题
答案
第十一章 极数训练
六、自测题答案
a,20,s 1. (1); (2); (3)收敛;(4)?1;(5)1; s,u,u12
,,11112nn2n(6)或;(7)( (π,3)[1,(,1)]xx,,2(2)!n2n!0n0n,,
2. (1) B; (2) B; (3) D; (4) C; (5) B; (6) D(
,,1111,,,,3. (1) ,又收敛,由比较判别法可知收敛; ,ln1,,ln1,,,,,,2222nnnn,,,,nn1,1,
πsin,1u11nn,1(2) ,所以由比值判别法可知收敛( lim,lim,,,1,πn,,n,,πnu22n,1n2sinsinnn,1
,,,,n11,n,1n,11n, 4. (1) u,(,1),则,因此,而发散,故uu,n,,,,nn222nnnnnn1n1,1nn,1,,
发散(
,111nnn,,,,但,由于是交错级数,且满足莱布尼茨定理,故收u(1)(1),(1)n,2nnnn1,
n,,,(,1)1,nn敛;而也收敛(即收敛(因此条件收敛( u(,1),,,n22nn,nn11n,1,
ππn,,,uarctan(2) u(1)arctan,( nn22nn
πarctan,,21πnlim,π因为,由比较判别法可知与同时收敛,故arctan,,22n,,1nnnn,1,12n
,πn绝对收敛( ,(1)arctan,2nn1,
n,an,3122n2n,,limlimR,((1)5,故级数的收敛半径为,从x,n,,n,,,a2(n2)22,2nn,1n,1
,,22,,,,而收敛区间为; ,,22,,
,annlim,1,1,x,1,11(2) ,故的收敛半径为,即时,级数au(u,x,1),nn,,an,1n1,
,n收敛,且收敛区间为( (,2,0)(x,1),n1,
n2n1,,,(,1)xn2n6. 设,则,( S(x),S(x),(,1)x,,2n,1n0n0,,
,,11nnnn2由基本展开式,,可知(而x,(,1,1)(,1)x,(,1)x,,x,(,1,1),,21,x1,xnn0,0,
xx1,, ,dx,arctanx,x,(,1,1)S(x)S(0)S(x)dx,,2,,001,x
所以 ( S(x),arctanx,x,(,1,1)
n2n2n2n1,,,,(,1)x(,1)x因此( ,x,,x,S(x),x,arctanx,x,(,1,1),,2n,12n,1n0n0,,
7. (1) ( f(x),ln(2,2x),ln2(1,x),ln2,ln(1,x)
,1nn,1, 由基本展式 ln(1,x),(,1)x,x,(,1,1),n,1n,0
,1nn,1可得 ( f(x),ln2,(,1)x,x,(,1,1),n,1n,0
1111f(x),,,, (2) ( x,11,x2,(x,1)21,2
,1nn 由基本展式 ,(,1)x,x,(,1,1),,1,xn0,
n,111,1,1xx,,n(),,,(,1),,1,,1,fx可得 ,,,,1x2222,,0,n1,2
n,(,1)n因此 ( f(x),(x,1),,1,x,3,1n,20n,