（I）数f?x?求导可得f'?x??cosx?xsinx?cosx??xsinx?x?0?,令f'?x??0可得
x?k??k?N*?,当x??2k?,?2k?1????k?N*?时,sinx?0.此时f'?x??0;
当x??2k?1??,?2k?2?????k?N*?时,sinx?0,此时f'?x??0, ???k?N*?,
故函数f?x?的单调递减区间为2k?,?2k?1??单调递增区间为
??2k?1??,?2k?2????k?N*?.
????x?,所以, ?01?22??(II)由(1)可知函数f?x?在区间?0,??上单调递减,又f?当n?N*时,因为f?n??f?1??n?1????????nn?1??0,且函数n??1????1??n?1???1???f?x?的图像是连续不断的,所以f?x?在区间?n?,?n?1???内至少存在一个零点,又f?x?在区间n?,?n?1??上是单调的,故n??xn?1??n?1??,因此, 当n?1时,
??142??; 22x1?31112; ??4?1???2x12x2?23当n?2时,当n?3时,
111+2?2?2x1x2x311?1?2?2?4?1?2?xn??2?11?1?2?2?5??xn??1?2??2??n?1???1??? n?2n?1?????1111?2+2?2?x1x2x3?111+2?2?x12x2x3?11??1??5??1???2xn?2???2?1???1??????n?2n?1???1?1?626???2?, 2???n?1??311?2?2x1x2?12?. 2xn3综上所述,对一切的n?N*,