Calculus Euler's Constant

A. by taking f(x)=1/x, show that
ln(n+1 )< 1 + 1/2 + ...+ 1/n < 1 + ln(n) OR 0 < ln(n+1) - ln(n) < 1 + 1/2 + ... + 1/n - ln(n) < 1
thus the sequence a(n) = 1 + 1/2 + ... + 1/n - ln(n)

B. show that 1/(n+1) < integral 1/x dx (from n to n+1) = ln(n+1) - ln(n) and use the result to show that the sequence a(n) in part A is dereasing
since a decreasing sequence that is bounded from below converges the number a(n) defined in part A converge: 1 + 1/2 + ... + 1/n - ln(n) -> y
the number y whose value is 0.5772 is called euler constant

step by step process please and thanks

A. We'll use integration and Left-Right Sums for the function f(x) = 1/x

Observe that 1/x is a strictly decreasing function in its domain.
Therefore, for any interval [a,b] in its domain and any partition Pn: a = t0 < t1 < ... < tn = b of [a,b], the left sum of f for this partition: RS(f,Pn) will be greater than the integral of f over [a,b].

So, as an integration interval we consider the interval [1, n+1]
And as a partition Pn we consider the points 1, 2, 3, ... n+1

As such: Integral[1 , n+1] f(x)dx = ln(n+1) , by definition of the logarithmic function

And: LS(f, Pn) = 1/1 * (2-1) + 1/2 * (3-2) + 1/3 * (4-3) + .... + 1/n * (n+1 - n) =

= 1/1 + 1/2 + 1/3 + ... + 1/n

And since, as explained above, Integral[1 , n+1] f(x)dx < LS(f, Pn) we have that:

ln(n+1) < 1/1 + 1/2 + 1/3 + ... + 1/n

Now let's partition in [1, n] and use the right sum, which will be smaller than f's integral:

RS(f, Pn-1) < Integral [1 to n] f(x)dx or

1/2 * (2-1) + 1/3 * (3-2) + ... + 1/n (n - (n-1)) < ln(n) <=>

1/2 + 1/3 + 1/4 + ... + 1/n < ln(n) <=>

1 + 1/2 + 1/3 + ... + 1/n < 1 + ln(n)

So we indeed get the double inequality:

ln(n+1 )< 1 + 1/2 + ...+ 1/n < 1 + ln(n) and subtracting ln(n) we get:

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