Question

已知數(shù)列{a_n},a₁=,且a_n+1=(1+)a_n+(n≥2,n∈N^*),b_n=(n∈N^*).

(1)當(dāng)n≥2時(shí),求證:a_n≥2;

(2)求證:當(dāng)x＞0時(shí),ln(1+x)＜x,且b_n＜e;

(3)在(2)的條件下,求證:a_n≤e³.

Answer 1

證明:(1)用數(shù)學(xué)歸納法證明:

①當(dāng)n=2時(shí),有a₂=(1+)a₁+1=2,a_n≥2成立.

②假設(shè)n=k時(shí),有a_k≥2,則當(dāng)n=k+1時(shí),有a_k+1=(1+)a_k+,

由a_k≥2,得a_k+1=(1+)a_k+≥2++＞2,所以a_k+1≥2.

由①②可知:當(dāng)n≥2時(shí),有a_n≥2成立.

(2)要證明b_n＜e成立,只需證＜e,

即ln(1+n)＜n,

當(dāng)x＞0時(shí),考察函數(shù)f(x)=ln(x+1)-x,有f′(x)=,易知f′(x)＜0,

∴f(x)=ln(x+1)-x在(0,+∞)上是單調(diào)遞減函數(shù).

∴f(x)＜f(0)=0.

則有l(wèi)n(x+1)-x＜0,∴l(xiāng)n(x+1)＜x成立,此時(shí)有l(wèi)n(n+1)＜n,則有＜e得證,∴b_n＜e.

(3)∵a_n≥2,∴1≤.∴a_n+1=(1+)a_n+≤(1+)a_n+·=(1++)a_n.

∴a_n+1≤(1++)a_n.

∴l(xiāng)na_n+1≤ln(1++)+lna_n.

則有l(wèi)na_n+1-lna_n≤ln(1++).

由(2)知,當(dāng)x＞0時(shí),有l(wèi)n(x+1)＜x成立,則有l(wèi)n(1++)≤+,

lna_n+1-lna_n≤+,

∴有l(wèi)na₂-lna₁≤+,lna₃-lna₂≤+,

……

lna_n-lna_n-1≤+.

各式相加得

lna_n-lna₁≤(++…+)+［1++++…+］.

∵++…+=1-,   

且1+++…+＜1+(1)+()+…+()=2,

∴l(xiāng)na_n-lna₁≤3＜3,

即有l(wèi)na_n≤lna₁+3=ln+3=lne³.

∴a_n≤e³.