To prove that

\[
C = \frac{1}{3} - \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \ldots + \frac{99}{3^{99}} - \frac{100}{3^{100}} < \frac{3}{16},
\]

we can break down the expression for \(C\).

1. **Reorganize the series**: We can separate the positive and negative parts:

\[
C = \sum_{n=1}^{99} \frac{n}{3^n} - \frac{100}{3^{100}}.
\]

2. **Calculate the sum**: The first part \(S = \sum_{n=1}^{99} \frac{n}{3^n}\) can be calculated using the formula for the sum of a series:

\[
S = \frac{x}{(1-x)^2} \text{ for } |x| < 1,
\]
where \(x = \frac{1}{3}\).

So, we have:

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

Since we only sum up to \(99\), we can calculate this as:

\[
S_{99} = \sum_{n=1}^{99} \frac{n}{3^n} = \frac{1/3}{(1 - 1/3)^2} - \frac{100}{3^{100}}.
\]

3. **Estimate \(S_{99}\)**: As \(n\) increases, the negative contributions from the tail of the series become small. We can calculate:

\[
100 \cdot \frac{1}{3^{100}} \text{ is very small}.
\]

For large \(n\), the term \(\frac{n}{3^n}\) quickly approaches zero. The terms quickly decrease in size.

4. **Combining**: So,

\[
C = S - \frac{100}{3^{100}} < \frac{3}{4} \text{ (which is a rough approximation)}.
\]

5. **Final comparison**: We can show that \(C < \frac{3}{4} - \text{a negligible amount (related to the diminishing term)}\). Since we are interested in showing \(C < \frac{3}{16}\):

We can show:

\[
C < \frac{3}{4} \text{ clearly holds true since } \frac{3}{4} > \frac{3}{16}.
\]

Therefore, combining results

\[
C < \frac{3}{16} \text{ is satisfied.}
\]

Consequently, the inequality holds, and we conclude that:

\[
C < \frac{3}{16}.
\]