TITLE: If $A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$ Then $\lfloor A \rfloor\;\;,$
QUESTION [6 upvotes]: If $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$$ Then $\lfloor A \rfloor\;\;,$ Where $\lfloor x \rfloor $ Represent floor fiunction of $x$

My Try:: Using $\bf{A.M\geq H.M\;,}$ We get
$$\frac{1980+1891+1982+....+2012}{33}>\frac{33}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}$$
So $$\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}<\frac{1980+1981+....+2012}{(33)^2}=\frac{1996}{33}\approx 60.5<61$$
Now how can i prove that the above expression $A$ is $>60$
Help me, Thanks

REPLY [7 votes]: HINT: There are $33$ terms in the denominator. The smallest is $\frac1{2012}$, and the largest is $\frac1{1980}$, so if $d$ is the denominator, then
$$\frac{33}{2012}\le d\le\frac{33}{1980}\;.$$
What does this tell you about $\frac{1}d$?