Chứng minh A = 1/3 - 2/3^2 + 3/3^3 - 4/3^4 + 5/3^5 - ... + 99/3^99 < 10/9

To prove that

\[
A = \sum_{n=1}^{99} \frac{(-1)^{n-1} n}{3^n} < \frac{10}{9},
\]

we can utilize the properties of series and inequalities.

### Step 1: Analyze the Series

We start with the series:

\[
A = \frac{1}{3} - \frac{2}{3^2} + \frac{3}{3^3} - \frac{4}{3^4} + \ldots + \frac{99}{3^{99}}.
\]

This is an alternating series with the general term

\[
a_n = \frac{n}{3^n}.
\]

### Step 2: Use Absolute Convergence

First, let's analyze the absolute series:

\[
\sum_{n=1}^{\infty} \left| a_n \right| = \sum_{n=1}^{\infty} \frac{n}{3^n}.
\]

This series can be evaluated by recognizing it as a power series. The formula for the sum of a series of the form

\[
\sum_{n=1}^{\infty} nx^n
\]

is given by

\[
\frac{x}{(1-x)^2} \quad \text{for}\ |x| < 1.
\]

In our case, we set \( x = \frac{1}{3} \):

\[
\sum_{n=1}^{\infty} n \left(\frac{1}{3}\right)^n = \frac{\frac{1}{3}}{(1 - \frac{1}{3})^2} = \frac{\frac{1}{3}}{\left(\frac{2}{3}\right)^2} = \frac{1/3}{4/9} = \frac{9}{12} = \frac{3}{4}.
\]

Thus,

\[
\sum_{n=1}^{\infty} \frac{n}{3^n} = \frac{3}{4}.
\]

### Step 3: Applying the Alternating Series Test

According to the alternating series test, since \( a_n \) is positive, monotonically decreasing, and converges to 0 as \( n \to \infty \), the series \( A \) converges.

### Step 4: Estimate the Remainder

Using the properties of alternating series, the error from truncating the infinite series at the \( n \)-th term is less than the absolute value of the next term:

\[
\left| R_n \right| = \left| \frac{100}{3^{100}} \right| < \frac{100}{3^{100}}.
\]

So,

\[
A \approx \sum_{n=1}^{\infty} \frac{(-1)^{n-1} n}{3^n} < \frac{3}{4} + \frac{100}{3^{100}} < \frac{3}{4} + \epsilon \ \text{(for very small } \epsilon).
\]

### Step 5: Final Comparison

Now just verify if \( \frac{3}{4} < \frac{10}{9} \):

\[
\frac{3}{4} = 0.75 \quad \text{and} \quad \frac{10}{9} \approx 1.1111,
\]

which is true.

### Conclusion

Thus, we have shown that:

\[
A < \frac{10}{9}.
\]

This completes our proof.