Wait, what?

a) It is sufficient to show that the following sequence $\frac{n!}{n}, \frac{n!}{n-1}, \frac{n!}{n-2} ... \frac{n!}{1}$ is in Harmonic Progression composed of increasing positive integers. b) For this we show that the difference of the corresponding A.P if such an infinite HP exists is $ < 0$. $a_n = \frac{a_1}{1+(n-1)da_1}$. Then we show that for a sufficiently large n the denominator of the nth term becomes less than 0 and hence $a_n < 0$ a contradiction.

If possible, let a1, a2, ... be an infinite harmonic sequence of positive integers. Then, 1/a1, 1/a2, .... is in AP. Let the common difference be d. 1/a2=(1/a1) + d = (1+d(a1))/a1 =1/(a1)/(1+d(a1)) Since a2 is an integer, 1+d(a1) divides a1 1/a3= (1+2d(a1))/a1=1/(a1)/(1+2d(a1)) Since a3 is an integer, 1+2d(a1) divides a1 Similarly, all numbers of the form 1+nd(a1) divide a1. This implies an infinite number of numbers divide a1. But, this is not possible given that a1 is a finite positive integer.

inmo2017 wrote: If possible, let a1, a2, ... be an infinite harmonic sequence of positive integers. Then, 1/a1, 1/a2, .... is in AP. Let the common difference be d. 1/a2=(1/a1) + d = (1+d(a1))/a1 =1/(a1)/(1+d(a1)) Since a2 is an integer, 1+d(a1) divides a1 1/a3= (1+2d(a1))/a1=1/(a1)/(1+2d(a1)) Since a3 is an integer, 1+2d(a1) divides a1 Similarly, all numbers of the form 1+nd(a1) divide a1. This implies an infinite number of numbers divide a1. But, this is not possible given that a1 is a finite positive integer. but $d$ is not necessarily an integer

b} if there exist a strictly increasing infinite sequence of positive integers which is in H.P, then the reciprocals of these terms,must be in A.P.{strictly decreasing} but there do not exist strictly decreasing A.P of positive integers ,because going on decreasing they will become negative

Here might be another way of doing the infinite part. Let $a_1,a_2,a_3,\dots$ be the infinite sequence of positive integers in harmonic progression such that $a_i>a_j$ if $i>j$. Then we have $\frac{1}{a_1}+\frac{1}{a_3}=\frac{2}{a_2},$ $\frac{1}{a_2}+\frac{1}{a_4}=\frac{2}{a_3},$ $\frac{1}{a_3}+\frac{1}{a_5}=\frac{2}{a_4}$ and so on. Adding these equations give $\frac{1}{a_1}+\frac{1}{a_2}+2\sum^{\infty}_{k=3}\frac{1}{a_k}=\frac{2}{a_2}+2\sum^{\infty}_{k=3}\frac{1}{a_k}$, which means $a_1=a_2$, contradiction.