Question 1085009
<pre>
Here's your second one:

(ii) The general term of {b<sub>n</sub>} is given by

        {{{b[n]=sqrt(1+n^2)-n}}}

         Prove that {{{b[n]>=b[n+1]}}}

Assume that it is false for contradiction, that is we assume:

{{{b[n]<b[n+1]}}}

{{{sqrt(1+n^2)-n<sqrt(1+(n+1)^2)-(n+1)}}}

{{{sqrt(1+n^2)-n<sqrt(1+(n+1)^2)-n-1}}}

{{{sqrt(1+n^2)<sqrt(1+n^2+2n+1)-1}}}

{{{sqrt(1+n^2)<sqrt(n^2+2n+2)-1}}}

{{{sqrt(1+n^2)+1<sqrt(n^2+2n+2)}}}

Since both sides are positive we can square both
sides and retain the inequality <

{{{(1+n^2)+2sqrt(1+n^2)+1<n^2+2n+2}}}

{{{1+n^2+2sqrt(1+n^2)+1<n^2+2n+2}}}

{{{2+n^2+2sqrt(1+n^2)<n^2+2n+2}}}

Isolate the radical term on the left:

{{{2sqrt(1+n^2)<2n}}}

Divide both sides by 2

{{{sqrt(1+n^2)<n}}}

Square both sides

{{{1+n^2<n^2}}}

{{{1<0}}}

A contradiction, so the inequality is proved.

{{{b[n]>=b[n+1]}}}

Edwin</pre>