1, n is equal to the minimum value that meets the conditions, indicating that it is established at this time. Generally speaking, what we write is obviously established without proof.
2. Assuming that n=k is true, it is proved that n=k+ 1 The difficulty lies in this step. (Denominators are generally calibrated, and denominators with root signs are physical and chemical. )
3, summary, the conclusion is established, generally as long as it is clearly established.
The greater than sign of this question should be the less than sign.
When n = 1, 1
If n=k- 1, it holds.
that is
1+
1/√2
+ 1/√3
+ ...
+ 1/√(k
- 1)& lt； 2√(k- 1)
Then when n=k,
1+
1/√2
+ 1/√3
+ ......
+ 1/√( k- 1)+ 1/√k & lt； 2 √ (k-1)+1√ k If there is 2 √ (k-1)+1√ k.
- 1)=2(√k-√(k
- 1)=2/[(√k+√(k
-1)], that is, as long as √(k
- 1 & lt； √k, and this is obviously. So 1+
1/√2
+ 1/√3
+ ......
+ 1/√n
& gt2√n