We can start by finding an upper and lower bound for the sum using integrals.

Lower bound:
We can approximate the sum as an integral:

∫₁¹₀₀ 1/x(x+1) dx

Using partial fractions, we can rewrite this as:

∫₁¹₀₀ (1/x - 1/(x+1)) dx

Evaluating the integral:

[ln(x) - ln(x+1)]₁¹₀₀ = ln(1) - ln(2) + ln(2) - ln(3) + ... + ln(99) - ln(100)

Simplifying:

ln(1/100) = -ln(100)

So the lower bound for A is:

A > -ln(100)

Upper bound:
We can use the same integral approximation, but this time we'll use a different method to evaluate it.

∫₁¹₀₀ 1/x(x+1) dx

We can use the fact that 1/x(x+1) < 1/x² for all x > 0.

So:

∫₁¹₀₀ 1/x(x+1) dx < ∫₁¹₀₀ 1/x² dx

Evaluating the integral:

[-1/x]₁¹₀₀ = 1/100 - 1/1

Simplifying:

1/100 - 1 > A

So the upper bound for A is:

A < 99/100

Putting it all together:

-ln(100) < A < 99/100

We can simplify this by using ln(100) = 2ln(10):

-2ln(10) < A < 99/100

And we can convert the upper bound to a common denominator:

-2ln(10) < A < 495/500

Finally, we can convert the lower bound to a common denominator:

-10ln(10) < A < 495/500

So:

-10ln(10) < A < 99/100

And in decimal form:

-2.3026 < A < 0.99