第七章 无穷级数 一.?选择题? 1.?设a为常数,?则级数 ? ￥ = ú ? ù ê ? é -?1? 2? 1?sin?n? n?n?na?(A)?绝对收敛.???(B)?发散.??(C)?条件收敛.??(D)?敛散性与a取值有关.? 解. ? ￥ =1? 2?sin?n? n?na 绝对收敛, ? ￥ =1?1?n? n?发散,?所以 ? ￥ = ú ? ù ê ? é -?1? 2? 1?sin?n? n?n?na 发散.?(B)是答案? 2.?设? )?1?1?ln(?)?1?(? n?u? n?n + - =? ,?则? (A) ? ￥ =1?n? n?u?与 ? ￥ =1?2?n? n?u?都收敛.??????(B) ? ￥ =1?n? n?u?与 ? ￥ =1?2?n? n?u?都发散.? (C) ? ￥ =1?n? n?u?收敛,?而 ? ￥ =1?2?n? n?u?发散.???(D) ? ￥ =1?n? n?u?发散, ? ￥ =1?2?n? n?u?收敛.? 解?.?由莱布尼兹判别法 ? ￥ =1?n? n?u?收敛?, ? ? ￥ = ￥ = + =?1? 2?1?2? )?1?1?(?ln?n?n? n? n?u .?因为? 1?)?1?1?(?ln?lim?2 = + ￥ ?? n?n?n? , ? ￥ =1?1?n? n?发散,?所以 ? ￥ =1?2?n? n?u?发散.?(C)是答案.? 3.?设函数? 1?0?,?)?(? 2 < ￡ =? x?x?x?f? ,?而 ? ￥ = +￥ < < -￥ =?1? ,?sin?)?(? n? n? x?x?n?b?x?s p? .?其中 ò = =? 1?0? )?,?2?,?1?(?,?sin?)?(?2? Ln?xdx?n?x?f?b?n p? ,?则? )2?1?(-?s?等于? (A)? 2?1 -?,?(B)? 4?1 -?,????(C)?4?1?,???????(D)?2?1? 解. ? +￥ < < -￥ =? x?x?n?b?x?s? n? ,?sin?)?( p 是? 1?0?,?)?(? 2 < ￡ =? x?x?x?f? 进行奇展拓后展成的富氏级 数.?所以? )2?1?(-?s? =? 4?1?)2?1(?)2?1( - = - = -? f?s? .?(B)是答案.? 4.?设 ? ￥ = -?1? )?1?(?n? n?n?a?条件收敛,?则? (A) ? ￥ =1?n? n?a?收敛,?????(B) ? ￥ =1?n? n?a?发散,???(C) ? ￥ = + -?1? 1?)?(?n? n?n? a?a? 收敛,? (D) ? ￥ =1?2?n? n?a?和 ? ￥ = +?1? 1?2?n? n?a?都收敛.? 解.?因为 ? ￥ = -?1? )?1?(?n? n?n?a?条件收敛,?所以? 0?lim = ￥ ?? n?n? a .?对于(C), ? = + + - = - =?nk? n?k?k?n? a?a?a?a?s 1? 1?1?1?)?(?所以? 1?1?1? )?(?lim?lim? a?a?a?s? n?n?n?n = - = + ￥ ? ￥ ?? .?(C)是答案.? 5.?设级数 ? ￥ =1?n? n?u?收敛,?则必定收敛的级数为? (A) ? ￥ = -?1? )?1?(?n? n?n?n?u (B) ? ￥ =1?2?n? n?u(C) ? ￥ = - -?1? 2?1?2? )?(?n? n?n? u?u (D) ? ￥ = - +?1? 1?)?(?n? n?n? u?u? 解. ? ￥ =1?n? n?u?收敛,?所以 ? ￥ = -?1? 1?n? n?u?收敛.?收敛级数的和收敛.?所以(D)是答案.?对于(C)有以下反 例: ? ? ￥ = - ￥ = - =?1? 1?1? 1?)?1?(?n? n?n? n? n?u , ? ? ￥ = ￥ = - - =?1?1? 1?2? 1?2?1?n?n? n? n?u , ? ? ￥ = ￥ = - =?1?1?2? 2?1?n?n? n? n?u .?所以 ? ? ￥ = ￥ = - = -? 1?1? 2?1?2? 1?)?(? n?n? n?n? n?u?u? 发散.? 6.?若 ? ￥ = -?1? )?1?(?n? n?n?x?a? 在? 2 - =?x?处收敛,?则此级数在? 1 - =?x?处? (A)?条件收敛,????(B)?绝对收敛,????(C)?发散,?(D)?收敛性不确定.?解.?因为在? 2 - =?x?收敛,?所以收敛半径大于?2.?幂级数在收敛半径内的任何点都绝对收敛.? (B)是答案.?7.?设幂级数 ? ￥ =1?n? n?n?x?a?的收敛半径为?3,?则幂级数 ? ￥ = + -?1? 1?)?1?(?n? n?n?x?na?的必定收敛的区间为?(A)?(－2,?4)??????(B)?[－2,?4]?????(C)?(－3,?3)?????(D)?(－4,?2)? 解. = ÷ ? ? ? è ? ? ￥ =? '?1?n? n?n?x?a ? ￥ = - =?1? 1?n? n?n?x?na ? ￥ = +?1? 1?2?n? n?n?x?na?x? 和 ? ￥ =1?n? n?n?x?a?有相同收敛半径.?所以?3?|?1?| < -?x? ,? 4?2 < < -? x?在(－2,?4)中级数一定收敛,?在端点级数不一定收敛.?所以答案为(A).? 二.?判断下列级数的敛散性:?1. ? ￥ = +?1? 1?sin?)?2?ln(?1?n? n?n? 解.?因为? 1?ln?1?1?sin?)?2?ln(?1?lim = + ￥ ?? n?n? n?n?n? ,?所以 ? ￥ = +?1? 1?sin?)?2?ln(?1?n? n?n? 和 ? ￥ =1? ln?1?n? n?n?有相同的敛散性.?又 因为 ò ￥ +?2? ln?1?dx?x?x? 发散,?由积分判别法知 ? ￥ =1? ln?1?n? n?n?发散.?所以原级数发散.? 2.? )?0?(?)?1?)(?)(?1?(? 1?1 1 + + + - + ? ￥ =? a?n?a?n?a?n?a?n? 解.?因为? 1?1? )?1?)(?)(?1?(? 1?lim? 3 = + + + - + ￥ ?? n? n?a?n?a?n?a?n? ,?所以? )?0?(?)?1?)(?)(?1?(? 1?1 1 + + + - + ? ￥ =? a?n?a?n?a?n?a?n? 和 ? ￥ =1? 3?1?n? n?有相同的敛散性. ? ￥ =1? 3?1?n? n?收敛,?所以原级数收敛.? 3. ? ￥ =1? !?3?n? n?n?n?n? 解.? 1?3?!?3?)?1?(? )!?1?(?3?lim?lim? 1?1?1 > = + + = + + ￥ ? + ￥ ?? e?n?n?n?n?u?u? n?n? n?n?n?n?n?n? ,?所以级数发散.? 4. ? ￥ = +?1? 2? )?/?1?(?n? n?n?n?n? 解.? 1?0?/?1?)?(?lim?)?/?1?(?lim? 2?2 < = + = + ￥ ? ￥ ?? n?n?n?n?n?n? n?n?n? n?n? ,?所以级数收敛.? 5. ? ￥ =1? 2?)!?2?(?)?!?(?n? n?n? 解.? 1?4?1?)?1?2?)(?2?2?(? )?1?(?lim?)!?2?(?!?!?)!?2?2?(? )!?1?(?)!?1?(?lim?lim? 2?1 < = + + + = + + + = ￥ ? ￥ ? + ￥ ?? n?n?n?n?n?n?n?n?n?u?u? n?n?n?n?n? ， 所以级数收敛.? 6. ? ￥ = + -?1? !?)2?2?(? !?)1?2?(?n? n?n? 解.?拉阿伯判别法: r = - + ￥ ?? )?1?(?lim?1?n?n?n? u?u?n?设? ,? 发散 收敛，? 1?1 < > r r? .? 2?3?1?2?3?lim?1?!?)4?2?(? !?)1?2?(? !?)2?2?(? !?)1?2?(?lim?)?1(?lim?1 = + = ÷ ÷ ÷ ÷ ? ? ? ? ? ? è ? - + + + - = - ￥ ? ￥ ? + ￥ ?? n?n?n?n?n?n?n?u?u?n? n?n?n?n?n? >?1,?所以级数收敛. 7. ? ￥ = - - +?1? )?1?1?(?!?1?n? n?n?n? 解.? 1?0?)?2?)(?1?(? 1?1?lim?)?1?1?(?!?1? )?2?(?)!?1?(?1?lim?lim?1 < = + + + - + + = - - + - + + = ￥ ? ￥ ? + ￥ ?? n?n?n? n?n?n?n?n? n?n?n?u?u? n?n?n?n?n? ,?级数收 敛.?8. ? ￥ = + - + +?1? 2?1?2? 1? )?1ln?2?(?n? n?n?n?n? ? 解.? 1?2?1?)?1?ln?2?(?lim?)?1?ln?2?(?lim? 2?1?2?2?2?1?2? 1 < = + + = + + + - ￥ ? + - ￥ ?? n? n?n?n? n?n?n? n?n?n?n?n?? ,?级数收敛.? 9. ? ￥ = -?1? )?ln?1?(?n? n?n?n? 解.?考察极限? y?y?y?n?y?n?n?n? y?y?n?n? 1?0? )?ln?1?(?lim?1?1?)?ln?1?(?lim + = - + ? ￥ ? 令 令? y?y?y?u? y?1?)?ln?1?( + =? ,? y? y?y?y?y?u? ln?)?ln?1?ln(?ln - + =? 1? 1?ln?ln?1? 1?ln?lim?ln?)?ln?1?ln(?lim?lim?ln?ln?lim? 0?0?0?0 - - + + = - + = = + + + + ? ? ? ?? y?y?y?y?y? y?y?y?y?u?u? y?y?y?y? =? 0?ln?1? ln?1?ln?ln?1?ln?lim? 2?0 = + - - - - + + ?? y?y? y?y?y?y?y?y?y? 所以? 1?lim?0?0 = = + ?? e?u?y? ,?即原极限为?1.?原级数和 ? ￥ =1?1?n? n?有相同的敛散性.?原级数发散.? 10. ? ￥ = + +?1? 1?)?1?(?n? n?n?n?n?n?n? 解.? 0?1?)?1?1?(?lim?)?1?(?lim? 2?1?1 1 = + × = + ￥ ? + ￥ ?? n?n? n?n?n?n?n?n?n? n?? ??n?n?n? ,?级数发散.? 三.?判断下列级数的敛散性?1.? L + - + -? 4?1?3?1?2?1? a?a?a?a? )?0?( >?a 解.?因为? 0?lim 1 ￥ ?? n?n? u ,?级数发散.? 2. ? ￥ = - + + + -?1? 1?1?)?1?(? 1?)?1?(?n? n? n?n?n? 解.? 0?1?1?)?1?(? 1?lim = - + + + ￥ ?? n?n?n?n? ,?令? 1?1?)?1?(? 1?)?( - + + + =? x?x?x?x?f? 当?x?>?0?时,? 0?]?11?)?1[(? 1?1?2?1?)?1(?)?(?'? 2?2 < - + + - + + - =? x?x? x?x?x?f? ,?所以数列 t y ü ? í ì - + + +? 1?1?)?1(? 1n?n? n? 单减.?根 据莱布尼兹判别法级数收敛.? 因为? 1?1?1?1?)?1?(? 1?lim = - + + + ￥ ?? n?n?n?n?n? ,?而 ? ￥ =1?1?n? n?发散,?所以 ? ￥ = - + + +?1? 1?1?)?1?(? 1?n? n?n?n? 发散.?原级数条 件收敛.?3. ? ￥ = ÷ ? ? ? è ? + + -?1? 1?3? 1?2?)?1(?n? n?n? n?n? 解.?因为? 3?2?1?3?1?2?lim = ÷ ? ? ? è ? + + ￥ ??n? n?n? n?n? ,?所以 ? ￥ = ÷ ? ? ? è ? + +?1? 1?3?1?2?n? n?n?n?收敛,?原级数绝对收敛.? 4. ? ￥ = - × × + × × -?1? )?1?3?(?8?5?2? )?1?2?(?7?5?3?)?1?(?n? n? n?n?L L 解.?因为? 1?3?2?2?3?3?2?lim?)?1?3?(?5?2? )?1?2?(?5?3? )?2?3?)(?1?3?(?5?2? )?3?2?)(?1?2?(?5?3?lim?lim?1 < = + + = - × + × + - × + + × = ￥ ? ￥ ? + ￥ ?? n?n?n?n?n?n?n?n?u?u? n?n?n?n?n? L L L L 所以 ? ￥ = - × × + × ×?1? )?1?3?(?8?5?2? )?1?2?(?7?5?3?n? n?n?L L 收敛,?原级数绝对收敛.? 5. ? ￥ = - -?1? 1? 1?tan?)?1?(?n? n? n?n? 解. ￥ ??n?lim? n?n?n?n?1?1?tan?=1, ? ￥ =1? 1?n? n?n?收敛,?原级数绝对收敛.? 6. ? ￥ = +?1? )?sin(?n? n?n p p 解. ? ? ￥ = ￥ = - = +? 1?1? sin?)?1?(?)?sin(? n? n?n? n?n?n p p p? .? 因为 p p = ￥ ?? n?n?n? 1?sin?lim? ,?又因为 ? ￥ = -?1? 1?)?1?(?n? n?n?,?条件收敛,?所以原级数条件收敛.? 四.?1.设正项数列? }?{?n?a?单调下降,?且 ? ￥ = -?1? )?1?(?n? n?n?a?发散,?证明:?级数 ? ￥ = + -?1? 1?)?1?(?n? n?na?a?收敛.? 2.?设正项数列? }?{?n?a,?}?{?n?b?满足? 0?(?1?1 > 3 - + + d d?n?n?n?n? b?a?a?b? 为常数),?证明:?级数 ? ￥ =1?n? n?a?收敛.? 证明:?1.?因为正项数列? }?{?n?a?单调下降,?且 ? ￥ = -?1? )?1?(?n? n?n?a?发散,?由莱布尼兹判别法,?a?a? n?n = ￥ ??lim?存在,?且? 0 1?a? .?容易证明:? a?a?n?n > "?,? .(反设存在?N,?使得? a?a?N <?.?则? k?N?N?N? a?a?a?a + + > > > > L?1? ,?令 ￥ ??k? ,?得到? a?a?a?N 3 >? ,?矛盾).?所以? a?a?a?a?a?a?a?a? n?n?n?n?n?n?n? 1?1?1?1 + + + - < - = -? .?因为 ? ￥ = + -?1? 1?n? n?n?a?a?a? 收敛,?所以 ? ￥ = + -?1? 1?)?1?(?n? n?na?a?收敛.? 2.?考察数列? }?{? n?n?a?b ,?因为? 0?(?1?1 > 3 - + + d d?n?n?n?n? b?a?a?b? 为常数?),?所以?0? 1?1?1 > 3 - + + +? n?n?n?n?n? a?a?b?a?b d? ,?即该数列递减有下界,?于是? n?n?n? a?b ￥ ??lim?存在.?由此推出 ? ￥ = + + -?1? 1?1? )?(?n? n?n?n?n? a?b?a?b? 收敛. d? 1?1?1 + + + - <? n?n?n?n?n? ababa ,?所以级数 ? ￥ =1?n? n?a?收敛.? 五.?求下列级数的收敛域:?1. ? ￥ = - +?1? 2?3? )?1?)(?3?(?n? n?n? x?n? 解. ? ? ? ￥ = ￥ = ￥ = - + - = - +? 1? 2?3?1? 2?1? 2?3? )?1?(?)?1?(?3?)?1?)(?3?(? n? n?n? n?n?n? n?n? x?n?x?x?n? 第一个级数的收敛半径为?3?1?,?第二个级数的收敛半径为?1.?所以它们的共同收敛区域为? )3?1?1?,3?1?1?( + -? .?考察端点: 当? 3?1?1± =?x? 时,?得 ? ? ￥ = ￥ = +?1?3?1? 3?1?n? n?n? n?第一个级数发散,?第二个级数收敛.?所以该级数发散.?原 级数的收敛区域为? )3?1?1?,3?1?1?( + -? .? 2. ? ￥ = ++ -?1? 1?2?1?2?)?1?(?n? n?n? n?x? 解.? 1?|?|?1?2?|?|?lim? 2?1?2 < = + + ￥ ?? x?n?x?n? n?n? ,?于是? 1?|?| <?x? .? 当? 1 =?x?时,?得 ? ￥ = + -?1? 1?2?1?)?1?(?n? n? n? ,?收敛?当? 1 - =?x?时,?得 ? ￥ = + + -?1? 1? 1?2?1?)?1?(?n? n? n? ,?收敛.?于是原级 数的收敛区域为[－1,?1].? 3. ? ￥ = - -?1? 1?2?2?1?2?n? n?n? x?n? 解.? 2?|?|?1?2?|?|?|?|?2?1?2?lim? 2?1?2 < < = - - ￥ ?? x?x?x?n?n? n?n?n? ，? .?当? 2 ± =?x? 时,?得数项级数 ? ￥ = -?1? 2?1?2?n? n?及 ? ￥ = - -?1? 2?1?2?n? n? ,?通项都不趋于?0,?发散.?该级数的收敛区域为? )?2?,?2?(-? .? 4. ? ￥ = ÷ ? ? ? è ? +?1? 2?1?n? n?n?n? x?x? 解. ? ? ? ￥ = ￥ = ￥ = + = ÷ ? ? ? è ? +? 1?1?1? 2?1?2?1? n? n?n?n? n?n? n?n?n? x?x?x?x? 第一个级数的收敛区域(－1,?1)??第二个级数的收敛区域? 2?1?|?| >?x? .?所以公共收敛区域为?)?12?1(?)2?1?1?(? ， ， è - -? .? 5. ? ￥ = × -?1? 2?9?)?1?(?n? n?n?n?x? 解.? 1?9?|?1?|?9?|?1?|?lim? 2?2 < - = × - ￥ ?? x?n?x?n? n?n?n? .?当? 3?1 ± = -?x? 时得数项级数 ? ￥ =1?1?n? n?,?发散.?该级数的收 敛区域为(－2,?4).? 6. ? ￥ = -?1? )?5?(?n? n?n?x? 解.? 1?|?5?|?|?5?|?1?lim < - = - ￥ ?? x?x?n?n? n?n? .?当? 1?5 - = -?x? 时,?得 ? ￥ = -?1? )?1?(?n? n?n?收敛,?当? 1?5= -?x? 时, 得 ? ￥ =1?1?n? n?发散敛.?该级数的收敛区域为[4,?6).? 六.?求下列级数的和:?1. ? ￥ = + + +?0? )?7?3?)(?4?3?)(?1?3?(? 1?n? n?n?n?解. ? ? = = ú ? ù ê ? é + + + - + = + + +? nk?nk? k?k?k?k?k?k? 1?1? 7?3?1?4?3?2?1?3?1?18?1?)?7?3?)(?4?3?)(?1?3?(? 1? = ú ? ù ê ? é + + + - + + + + - + + - + + -? 7?3?1?4?3?2?1?3?1?13?1?10?2?7?1?10?1?7?2?4?1?7?1?4?2?1?18?1? n?n?n?L? =? 24?1?7?3?1?4?3?1?4?1?1?18?1 ? ú ? ù ê ? é + + + - -? n?n? .? 所以? 24?1?)?7?3?)(?4?3?)(?1?3?(? 1?0 = + + + ? ￥ =?n? n?n?n? 2. ? ￥ = +?1? )?(?1?n? m?n?n? 解. ÷ ? ? ? è ? + - = +? m?n?n?m?m?n?n? 1?1?1?)?(?1? .?令?n?充分大, ￥ ??n ? ? ? = ￥ ? = ￥ ? ￥ = ÷ ? ? ? è ? + - = + = +? nk?n?nk?n?n? m?k?k?m?m?k?k?m?n?n? 1?1?1? 1?1?lim?1?)?(?1?lim?)?(?1?= ÷ ? ? ? è ? + + + = ÷ ? ? ? è ? + - - + - + + + ￥ ?? m?m?m?n?n?m?m?n? 1?2?1?1?1?1?1?1?1?2?1?1?lim?1? L L L?3. ? ￥ = - - - -?1? 1?2?1? 1?2?)?1?(?n? n?n? n?x?解.? 1?|?|? 1?2?|?|?lim? 2?1?2 < = - - ￥ ?? x?n?x?n? n?n? 级数收敛,?所以收敛半径为?1.?当? 1 ± =?x?时都得到交错级数.?由 莱布尼兹判别法知收敛.?所以收敛区域为[－1,?1].令 =?)?(x?s ? ￥ = - - - -?1? 1?2?1? 1?2?)?1?(?n? n?n? n?x? . =?)?(?'?x?s 2?1? 2?2?1? 1?1?)?1?(? x?x?n? n?n + = - ? ￥ = - - 所以? x?dx?x?dx?x?s?x?s? x?x? arctan?1?1?)?(?'?)?(? 0? 2?0 = + = = ò ò? ,?[－1,?1].? 4. ? ￥ = +?1? )?1?(?n? n?x?n?n? 解.? 1?|?|?|?|?)?1(?lim < = + ￥ ?? x?x?n?n?n? n?n? 收敛.?当? 1 ± =?x?得 ? ￥ = +?1? )?1?(?n? n?n?及 ? ￥ = + -?1? )?1?(?)?1?(?n? n? n?n?都发散. 所以收敛区域为(－1,?1).? 3?'?2?1? '?1? 1?1?1? )?1?(?2?1?)?1?(?)?1?(? x?x?x?x?x?x?x?x?n?n?x?x?n?n? n? n? n?n?n? n + = ÷ ? ? ? è ? + = ÷ ? ? ? è ? + = + ? ? ? ￥ = ￥ = + - ￥ = 积分二次? ,(－1,?1)? 5. ? ￥ = +?1? 2?)?1(?n? n?n?n?x? 解.? 1?2?|?1?|?2?|?1?|?lim < + = + ￥ ?? x?n?x?n? n?n?n? ,?所以当? 1?3 < < -? x?时收敛.? 当? 1 =?x?时得数项级数 ? ￥ =1?1?n? n?,?发散??当? 3 - =?x?时得数项级数 ? ￥ = -?1? 1?)?1?(?n? n?n?,?收敛.?于是收敛区 域为[－3,?1). ò ò ? ò ? ? - - ￥ = - - ￥ = ￥ = - = ÷ ? ? ? è ? + = ÷ ? ? ? è ? + = +? x?x?n? n?x?n? n?n?n? n?n? dx?x?dx?x?dx?n?x?n?x? 1?1? 1? 1?'?1? 1?1? ?2?2?1?2?1?2?1?2?)?1(?2?)?1(? =? x?x - = - -? 1?2?ln?)?1?ln(?2?ln? ,??[－3,?1).? 6. ? ? ￥ = - ￥ = - - + +? 1? 1?1? 1?1? 2?)?1?(?2?)?1?(? n? n?n? n?n? n?n?x?n?n? 并求 解.? 2?|?|?,?1?2?|?|?|?|?2?)?1?(?lim? 1?1 < < = + - - ￥ ?? x?x?x?n?n?n? n?n?n? .?当? 2 ± =?x?时得到的数项级数发散,?所以收敛 区域为(－2,?2).? 3?'?2?'?1? 1?'?1? 1?1?1? 1?1? )?2?(?16?2?2?2?4?2?2? )?1(? x?x?x?x?x?x?n?n n? n?n? n?n?n? n?n - = ÷ ? ? ? è ? - = ÷ ÷ ? ? ? ? è ? ÷ ? ? ? è ? = ÷ ? ? ? è ? + ? ? ? ￥ = + ￥ = - + ￥ = - - 积分二次? ,?(－2,?2)? 所以? 16?)?1?2?(?16?2?)?1?(? 3?1? 1 = - = + ? ￥ = -?n? n?n?n? 七.?把下列级数展成x?的幂级数:?1.? )?2?1?ln(?)?(? 2?x?x?x?f - + = 解.? x?x?x?x?x?x?f - - + = - + - =? 1?1?1?2?2?2?1? 4?1?)?(?'? 2? = ? ? ? ￥ = ￥ = + ￥ = - - = - - ? 0? 0? 1?0? ]?1?2?)?1?[(?2?)?1?(?2? n? n? n?n?n?n?n? n?n?n? x?x?x? 上述级数的收敛半径为?2?1?.?所以 = - - = ò ? ￥ = +?x?n? n?n?n? dx?x?x?f? 0? 0? 1? ]?1?2?)?1?[(?)?(? n?n? n?n?n? n?n?n? x?n?n?x ? ? ￥ = - ￥ = + + - - = + - -? 1? 1?0? 1?1? ]?1?2?)?1?[(?1?]?1?2?)?1?[( 当? 2?1 - =?x? 时得数项级数 ? ￥ = - - -?1? 2?1?)?1?(?]?1?2?)?1?[(?n? n?n?n?n?n? .?因为 ? ￥ = -?1? 1?n? n?发散, ? ￥ = + -?1? 1? 2?1?)?1?(?n? n?n? n?收 敛,?所以 ? ￥ = - - -?1? 2?1?)?1?(?]?1?2?)?1?[(?n? n?n?n?n?n? 发散.? 当? 2?1 =?x?时得数项级数 ? ￥ = - -?1? 2?1?]?1?2?)?1?[(?n? n?n?n?n? .?因为 ? ￥ = -?1? 1?)?1?(?n? n?n?收敛, ? ￥ = -?1? 2?1?n? n?n?收敛,?所以 ? ￥ = - -?1? 2?1?]?1?2?)?1?[(?n? n?n?n?n? 收敛.?所以 = - - = ò ? ￥ = +?x?n? n?n?n? dx?x?x?f? 0? 0? 1? ]?1?2?)?1?[(?)?(? n?n? n?n?n? n?n?n? x?n?n?x ? ? ￥ = - ￥ = + + - - = + - -? 1? 1?0? 1?1? ]?1?2?)?1?[(?1?]?1?2?)?1?[(? 的收敛区域为? ]2?1,2?1?(-? .? 2.? )?2?1?)(?1?(? 3?)?(? x?x?x?f + - = 解. ? ? ￥ = ￥ = + - = - + + = + - =? 0? 1?2?)?1?(?2?1?1?2?1?2?)?2?1?)(?1?(? 3?)?(? n? n? n?n?n?n? x?x?x?x?x?x?x?f? = ? ￥ = + + -?0? 1? ]?1?2?)?1?[(?n? n?n?n? x? 以上级数在公共收敛区域? )2?1,2?1?(- 内收敛.? 当? 2?1 - =?x? 时得 ? ￥ = - +?0? ]?2?1?)?1?(?2?[?n? n?n? ,?发散?? 当? 2?1 =?x?时得 ? ￥ = + -?0? ]?2?1?2?)?1?[(?n? n?n? ,?发散.?所以? )?(x?f? = ? ￥ = + + -?0? 1? ]?1?2?)?1?[(?n? n?n?n? x?,? )2?1,2?1?(-? 3.? 2?2? 1?)?1?ln(?)?(? x?x?x?x?x?f + - + + = 解.? )?1?ln(?1?1?)?1?ln(?)?(?'? 2?2?2?2? x?x?x?x?x?x?x?x?x?f + + = + - + + + + =? n?n? x?n? n?x?x?x?f? 2?1?2?1?2?2? !? )?1?2?1?(?)?2?2?1?(?)?1?2?1?(?2?1?1?)?1?(?1?1?)?(?'?' ? ￥ = - + - - - - × - - × - + = + = + = L =? n?n? n?n?n? n? x?n?n?x?n? n? 2?1?2?1? !?2? !?)1?2?(?)?1(?1?! 2? )?12?(?1?2?4?1?2?2?1?2?1?)?1(?1 - - + = - + + × + × - + ? ? ￥ = ￥ = L 注意到? 1?)?0?(?0?)?0?(?' - = =? f?f? ，? .?将上式由?0到?x?积分二次,?得 ? ￥ = + + + - - + = +? 1? 2?2?2? )?12?(?!?)2?2?(? !?)1?2?(?)?1(?2?1?)?(? n? n?n? x?n?n?n?x?x?f? ,?即 ? ￥ = + + + - - + + - =? 1? 2?2?2? )?12?(?!?)2?2?(? !?)1?2?(?)?1(?2?1?)?(? n? n?n? x?n?n?n?x?x?f? 当? 1 ± =?x?时,?考察数项级数 ? ￥ = + + -?1? )?1?2?(?!?)2?2?(? !?)1?2?(?n? n?n?n? .?由拉阿伯判别法? 5?.?2?)?12?(? 1?10?1? 2?2?1 ? + - = ú ? ù ê ? é - +? n? n?n?a?a?n?n?n? 数项级数收敛.?所以 ? ￥ = + + + - - + + - =? 1? 2?2?2? )?1?2?(?!?)2?2?(? !?)1?2?(?)?1?(?2?1?)?(? n? n?n? x?n?n?n?x?x?f? ,???[－1,?1]? 八.?把下列级数在指定点展成幂级数:?1.? 2?3?1?)?(? 2 + + =? x?x?x?f? 在? 4?0 - =?x? 处 解.? 2?4?1?3?4?1?2?1?1?1?2?3?1?)?(? 2 - + - - + = + - + = + + =? x?x?x?x?x?x?x?f? = ? ￥ = + + + ÷ ? ? ? è ? - = + - × - + - ×? 0? 1?1? )?4?(?3?1?2?1?3?4?1?1?3?1?2?4?1?1?2?1? n? n?n?n? x?x?x? ,??(－6,－2)? 2.? x?x?f? lg?)?( =? ,?在? 1?0 =?x?处 解. ? ￥ = + + - - = - + = = =? 0? 1?1?)?1?(?)?1?(?10?ln?1?)?1?1?ln(?10?ln?1?ln?10?ln?1?lg?)?(? n? n?n? n?x?x?x?x?x?f? ,??(0,?2]? 九.?把下列函数分别展成正弦函数和余弦函数:?1.? 2?)?(? x?x?f =? ,?( p?2?0 < <?x? )? 解.?1.?偶展拓? 3?8?3?1?1? 3?2?0?3?2?0? 2?0 p p p p p = = = ò x?dx?x?a ú ú ? ù ê ê ? é - = = = ò ò ò p p p p p p p? 2?0?2?0?2?2?0? 2?2?0? 2? 2?sin?2?2?sin?2?2?sin?2?2?cos?1? dx?nx?x?nx?x?n?nx?d?x?n?dx?nx?x?a?n? = ú ú ? ù ê ê ? é - = = - ò ò ò p p p p p p p? 2?0?2?0?2?2?0?2?2?0? 2?cos?2?cos?8?2?cos?8?2?sin?4? dx?nx?nx?x?n?nx?xd?n?dx?nx?x?n? =? 2?2? 16?)?1?(?2?2?cos?2?8? n?n?n? n - = × p p p? .? 于是 ? ? ￥ = ￥ = - + = + = =? 1? 2?2?1?0?2? 2?cos?16?)?1?(?3?4?2?cos?2?)?(? n? n?n? n? nx?n?nx?a?a?x?x?f p? ,???[0,?2p]?2.?奇展拓 ú ú ? ù ê ê ? é - - = - = = ò ò ò p p p p p p p? 2?0?2?0?2?2?0? 2?2?0? 2? 2?cos?2?2?cos?2?2?cos?2?2?sin?1? dx?nx?x?nx?x?n?nx?d?x?n?dx?nx?x?b?n?= ò + - - p p p p? 2? 0?2?2? 2?sin?8?)?1(?4?2? nx?xd?n?n? n p p p p p p p? 2?0?3?1?2?0?2?0?2?1? 2?cos?16?8?)?1?(?2?sin?2?sin?8?8?)?1?(? nx?n?n?dx?nx?nx?x?n?n? n?n + - = ú ú ? ù ê ê ? é - + - = + + ò =? ]?1)?1[(?16?8?)?1(? 3?1 - - + - +? n?n? n?n p p? .?当? k?n?2 =? ,? k?k?b? k?k p p? 4?2?8?)?1(? 1?2?2 - = - = + 当? 1?2 + =?k?n? , ú ? ù ê ? é + - + = + - + = +? 3?2?3?1?2? )?12?(?4?1?2?8?)?12?(?32?1?2?8? k?k?k?k?bk p p p p ? ￥ = = =? 1?2? 2?sin?)?(? n? n? nx?b?x?x?f? ,???(0,?2p)? 2. ? í ì - =? x?x?x?f? 2?)?(? 2?1? 1?0 < < ￡ <?x?x? 解:?1.?偶展拓? 1?2?1?2?1?)?2?(?)?(? 2? 1?1?0?2?0?0 = + = - + = = ò ò ò dx?x?xdx?dx?x?f?a ò ò ò - + = =? 2? 1?1?0?2?0? 2?cos?)?2?(?2?cos?2?cos?)?(? dx?x?n?x?dx?x?n?x?dx?x?n?x?f?a?n p p p?=? ]?1)?1[(?2? 2?2 - -?n?n p 当? k?n?2 = 时,? 0? 2 =?k?a 当? 1?2 + =?k?n? 时,? 2?2?1?2? )?1?2?(? 4 p + - = +? k?a?k? 于是 ? ￥ = + + - =? 0? 2?2? )?1?2?(? 2?1?2?cos?4?2?1?)?(? k? k?x?k?x?f p p? ,????(1?<?x<?2)? 2.?奇展拓 ò ò ò - + = =? 2? 1?1?0?2?0? 2?sin?)?2?(?2?sin?2?sin?)?(? dx?x?n?x?dx?x?n?x?dx?x?n?x?f?b?n p p p? =? n?n? )?1(?8?2?2 - p 于是 ? ￥ = - =? 1? 2?2? 2?sin?)?1?(?8?)?(? n? n? n?x?n?x?f p p? ,????(1?<?x<?2)? 十.?把下列函数展成付氏级数?1.? |?|?cos?2?1?)?(? x?x?x?f + =? ,???[－p, p]? 解.?|x|为偶函数,?将它展拓成以?2p为周期的周期函数. p p p = = ò? 00? 2? xdx?a ò ò = = p p p p? 0?0? cos?2?cos?)?(?2? nxdx?x?nxdx?x?f?a?n? =? ]?1)?1[(?2?2 - -?n?n p 当? k?n?2 = 时,? 0? 2 =?k?a? 当? 1?2 + =?k?n? 时, p?2?1?2? )?1?2?(? 4 + - = +? k?a?k? 于是 ? ￥ = - - - - + = + =? 2? 2?)?1?2?(? )?1?2?cos(?4?cos?)?4?2?1(?2?|?|?cos?2?1?)?(? k? k?x?k?x?x?x?x?f p p p? ,???[－p, p]? 2. ? ? ? ? ? í ì - =?2?2?)?( p p?x?x?f p p p p p p < ￡ < ￡ - - < ￡ -? x?x?x?2? 2?2? 2? 解:? )?(x?f?为奇函数,?展拓成以?2p为周期的周期函数 ò ò ò + = = p p p p p p p? 2?2?0?0? sin??sin?2?sin?)?(?2? nxdx?nxdx?x?nxdx?x?f?b?n? =? n?n?n?n? )?1(?1?2?sin?2?2 - - p p 于是? nx?n?n?n?x?f? n? n?sin?)?1?(?1?2?sin?2?)?(? 1? 2 ? ￥ = ú ? ù ê ? é - - = p p = ? ￥ = ú ? ù ê ? é - -?1? sin?2?)?1?(?2?sin?1?1?2?n? n? nx?n?n?n p p p? ,????(－p, p)