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

2 Answers
Jul 15, 2018

Answer:

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#

Jul 15, 2018

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.