Question

Show that `1+1/sqrt2+cdots+ 1/sqrtn >=sqrt2(n-1)`, for `n >1`?

Answer 1

Below

Explanation:

To show that the inequality is true, you use mathematical induction

`1+1/sqrt2+...+1/sqrtn >=sqrt2(n-1)` for `n >1`

Step 1: Prove true for `n=2`
LHS=`1+1/sqrt2`
RHS=`sqrt2(2-1)=sqrt2`
Since `1+1/sqrt2 >sqrt2`, then `LHS > RHS`. Therefore, it is true for `n=2`

Step 2: Assume true for `n=k` where k is an integer and `k >1`

`1+1/sqrt2+...+1/sqrtk >=sqrt2(k-1)` --- (1)

Step 3: When `n=k+1`,
RTP: `1+1/sqrt2+...+1/sqrtk+1/sqrt(k+1) >=sqrt2(k+1-1)`
ie `0 >=sqrt2-(1+1/sqrt2+...+1/sqrtk+1/sqrt(k+1))`

RHS
=`sqrt2-(1+1/sqrt2+...+1/sqrtk+1/sqrt(k+1))`
`>=sqrt2-(sqrt2(k-1)+1/sqrt(k+1))` from (1) by assumption
=`sqrt2-sqrt2(k)+sqrt2-1/sqrt(k+1)`
=`2sqrt2-sqrt2(k)-1/sqrt(k+1)`

Since `k>1`, then `-1/sqrt(k+1)<0` and since `ksqrt2>=2sqrt2>0`, then `2sqrt2-ksqrt2<0` so `2sqrt2-sqrt2(k)-1/sqrt(k+1)=<0`
=LHS

Step 4: By proof of mathematical induction, this inequality is true for all integers `n` greater than `1`

Answer 2

The inequality as stated is false.

Eg, for `n = 3`:

`underbrace(1+1/sqrt2+ 1/sqrt3)_(approx 2.3) cancel( >=) underbrace(sqrt2(3-1))\_( approx 2.8)`

A contradiction.