If $A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}\;,$ Then $\lfloor A \rfloor\;\;,$

Question

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

Answer 1

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 $\frac{1}d$?

Answer 2

The elementary estimate

$$\frac{m}{n+m} < \sum_{k = 0}^{m-1} \frac{1}{n+k} < \frac{m}{n}$$

for $m > 1$ (we have equality on the right for $m = 1$) gives

$$\frac{1}{61} = \frac{33}{2013} < \sum_{k = 1980}^{2012} \frac{1}{k} < \frac{33}{1980} = \frac{1}{60},$$

whence $\lfloor A\rfloor = 60$.

Answer 3

Let $$A=\frac{1}{\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}}=\frac 1K$$ $$K=\frac{1}{1980}+\frac{1}{1981}+\frac{1}{1982}+........+\frac{1}{2012}<\frac{1}{1980}+\frac{1}{1980}+\frac{1}{1980}+........+\frac{1}{1980}=\frac{33}{1980}=\frac{1}{60}$$ $$A=\frac 1K >60$$