# show the string an decrease when an=1+1/2+...+1/n - ln(n+1/2) f(x)=1/(x+1)-ln(x+3/2) +ln(x+1/2) x>0

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You need to test if the given sequence decreases, such that:

`a_(n+1) - a_n < 0`

`a_(n+1) - a_n = 1 + 1/2 +...1/n + 1/(n+1) - ln(n + 1 + 1/2) - 1 - 1/2 -...1/n + ln(n + 1/2)`

Reducing duplicate members yields:

`a_(n+1) - a_n = 1 - ln(n + 3/2) + ln(n + 1/2)`

You need to notice that the problem provides the function `f(x) = 1 - ln(x + 3/2) + ln(x + 1/2)` and comparing `a_(n+1) - a_n` with the equation of function yields:

`a_(n+1) - a_n = f(n)`

You need to test if the function decreases, hence, you need to use its derivative, such that:

`f'(n) = (1 - ln(n + 3/2) + ln(n + 1/2))'`

`f'(n) = -1/(n + 3/2) + 1/(n + 1/2) => f'(n) = -2/(2n + 3) + 2/(2n + 1)`

`f'(n) = 2(-2n - 1 + 2n + 3)/((2n+1)(2n+3))`

`f'(n) = 4/((2n+1)(2n+3))`

Since the function `f'(n)` is positive for `n>0` , hence `f(n)` increases over `(0,oo)` .

`a_(n+1) - a_n = f(n) > 0 => a_(n+1) > a_n`

**Hence, testing if the given sequence a_n decreases, using derivative of function, yields that it increases over `(0,oo).` **

`a_n=1+1/2+1/3+..........+1/n-ln(n+1/2)`

`a_(n+1)=1+1/2+1/3+.......+1/n+1/(n+1)-ln(n+1+1/2)`

`a_(n+1)-a_n=1/(n+1)-ln(n+3/2)+ln(n+1/2)`

`=1/(n+1)-ln(2n+3)+ln(2n+1)`

`=1/(n+1)-ln((2n+3)/(2n+1))`

`=1/(n+1)-ln(1+1/(2n+1))`

`ln(1+1/(2n+1))>1`

`-ln(1+1/(2n+1))<-1`

so `1/(n+1)-ln(1+1/(2n+1))<1/(n+1)-1`

`1/(n+1)-ln(1+1/(2n+1))<(1-n-1)/(n+1)`

`1/(n+1)-ln(1+1/(2n+1))<-1/(n+1)<0`

`1/(n+1)-ln(1+1/(2n+1))<0`

`` Thus `a_n` is decreasing.

From above we can find realÂ function f(x) ,x>0 defined as,

`f(x)=1/(x+1)-ln(2x+3)+ln(2x+1)`

Differentiate fÂ w.r.t x ,we have

`f'(x)=-1/(x+1)^2-2/(2x+3)+2/(2x+1)`

`=-1/(x+1)^2-2((2x+1-2x-3)/(4x^2+8x+3))`

`={-(4x^2+8x+3)+4(x^2+2x+1)}/{(x+1)^2(4x^2+8x+3)}`

`=1/{(x+1)^2(4x^2+8x+3)}`

`f'(x)>0 ,x>0`

This means function f(x) is an increasing fuction for x>0 .