(1)∵函数f(x)=e x-ln(x+1),
∴f′(x)=e x-
1
x+1 ,
∴k=f′(0)=e 0-
1
0+1 =0,
f(0)=e 0-ln1=1,
∴曲线y=f(x)上一点(0,f(0))处的切线方程为:y-1=0.
(2)∵f′(x)=e x-
1
x+1 ,x>-1.
∴由f′(x)=e x-
1
x+1 =0,得x=0.
当x>0时,e>1,
1
x+1 <1,所以当x>0时,f′(x)>0;
当-1<x<0时,ex<1,
1
x+1 >1,所以当x<0时,f′(x)<0.
∴函数f(x)的减区间是(-1,0),增区间是(0,+∞).
(3)∵函数f(x)的减区间是(-1,0),增区间是(0,+∞),
∴当x=0时,f(x)取得最小值f(0)=1,∴f(x)≥1,
∴e x-ln(x+1)≥1,即e x≥ln(x+1)+1,
取x=
1
n ,则 e
1
n ≥ln(
1
n +1)+1=ln(n+1)-lnn+1,
于是e≥ln2-ln1+1,
e
1
2 ≥ln3-ln2+1,
e
1
3 ≥ln4-ln3+1,
…
e
1
n ≥ln(n+1)-lnn+1.
相加得,e+ e
1
2 + e
1
3 +…+ e
1
n ≥ln(n+1)+n.(n∈N*,e为常数).
故 e+ e
1
2 + e
1
3 +…+ e
1
n ≥ln(n+1)+n(n∈ N * ,e为常数) .