Cho \( A = \frac{2}{3^2} + \frac{2}{5^2} + \frac{2}{7^2} + \cdots + \frac{2}{2011^2} \). Chứng minh rằng \( A < \frac{1005}{2012} \)

To prove that

\[
A = \frac{2}{3^2} + \frac{2}{5^2} + \frac{2}{7^2} + \cdots + \frac{2}{2011^2} < \frac{1005}{2012},
\]

we can find an upper bound for the sum \( A \).

### Step 1: Rewrite the sum
First, notice that \( A \) is the sum of the inverses of the squares of the odd numbers from 3 to 2011. We can express this as:

\[
A = 2 \sum_{k=1}^{1005} \frac{1}{(2k + 1)^2},
\]

where \( 2k + 1 \) generates the odd numbers from 3 to 2011.

### Step 2: Understand the series
Let’s denote

\[
S = \sum_{k=1}^{1005} \frac{1}{(2k + 1)^2}.
\]

### Step 3: Find an upper bound for \( S \)
We can upper bound \( S \) by comparing it to an integral:

\[
S < \int_{1}^{1006} \frac{1}{(2x + 1)^2} \, dx.
\]

Calculating this integral, we have:

\[
\int \frac{1}{(2x + 1)^2} \, dx = -\frac{1}{2(2x + 1)} + C.
\]

Then, evaluating from 1 to 1006:

\[
\int_{1}^{1006} \frac{1}{(2x + 1)^2} \, dx = \left[-\frac{1}{2(2(1006) + 1)} \right] - \left[-\frac{1}{2(2(1) + 1)} \right]
\]

Calculating the values:

- At \( x = 1006 \):
\[
-\frac{1}{2(2013)} = -\frac{1}{4026}.
\]
- At \( x = 1 \):
\[
-\frac{1}{2(3)} = -\frac{1}{6}.
\]

So,

\[
\int_{1}^{1006} \frac{1}{(2x + 1)^2} \, dx = -\frac{1}{4026} + \frac{1}{6}.
\]

### Step 4: Simplify the calculation
Now, calculate \( \frac{1}{6} - \frac{1}{4026} \):
\[
\frac{1}{6} = \frac{671}{4026} \quad (\text{since } 6 \times 671 = 4026),
\]
thus,
\[
\frac{671}{4026} - \frac{1}{4026} = \frac{670}{4026} = \frac{335}{2013}.
\]

### Step 5: Find \( S \) and conclude
Therefore,
\[
S < \frac{335}{2013}.
\]

Thus, we have:
\[
A < 2S < 2 \cdot \frac{335}{2013} = \frac{670}{2013}.
\]

To compare with \( \frac{1005}{2012} \):

Calculating \( 670 \times 2012 \) versus \( 1005 \times 2013 \):

\[
670 \times 2012 = 134,5840,
\]

\[
1005 \times 2013 = 2,020,065.
\]

Since

\[
1345840 < 2020065,
\]

we conclude:

\[
\frac{670}{2013} < \frac{1005}{2012}.
\]

Thus,

\[
A < \frac{1005}{2012},
\]

proving the inequality is valid. Hence, we have shown that:

\[
A < \frac{1005}{2012}.
\]

Therefore,

\[
\boxed{A < \frac{1005}{2012}}.
\]