To evaluate the series

\[
Q = \sum_{n=1}^{2022} \frac{(-1)^{n-1} n}{5^n},
\]

we can look for a way to sum this series conveniently.

The series can be rewritten in a more manageable form. First, consider the function

\[
S(x) = \sum_{n=1}^{\infty} n x^n,
\]

where \( |x| < 1 \). The formula for this series is related to the derivative of the geometric series. Recall that

\[
\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x} \quad \text{for } |x| < 1.
\]

Taking the derivative with respect to \( x \):

\[
\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}.
\]

Multiplying by \( x \) gives:

\[
\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2}.
\]

Now, we want to consider \( S(-1/5) \):

\[
S\left(-\frac{1}{5}\right) = \sum_{n=1}^{\infty} n \left(-\frac{1}{5}\right)^n = \frac{-\frac{1}{5}}{\left(1 + \frac{1}{5}\right)^2} = \frac{-\frac{1}{5}}{\left(\frac{6}{5}\right)^2} = \frac{-\frac{1}{5}}{\frac{36}{25}} = -\frac{25}{180} = -\frac{5}{36}.
\]

Since we desire only the first 2022 terms of this series and a negative sign is accounted for, the partial sum can be determined using

\[
Q = S\left(-\frac{1}{5}\right) - \frac{2022}{5^{2022}} \quad \text{(contribution from } n=2022 \text{ omitted)}.
\]

However, due to the behavior of the series with alternating signs, we note:

1. Each term \( \frac{n}{5^n} \) gets smaller, contributing less and less as \( n \) increases.
2. The series converges rapidly due to the factor \( 5^n \) in the denominator.

So we have:

\[
Q = -\frac{5}{36} + \frac{2022}{5^{2022}}.
\]

The critical part here is evaluating the term \( \frac{2022}{5^{2022}} \). It's evident that for relatively small \( n \), \( Q \) will be dominated by the sum of the first few terms, with additional terms rapidly decreasing. The term \( \frac{2022}{5^{2022}} \) will be very small compared to \( -\frac{5}{36} \).

Now, since \( Q < -\frac{5}{36} + \text{(a small positive term)} \), we see that

\[
Q < 0 \quad \text{and definitely} \quad Q < -\frac{5}{36} \text{ for all terms up to } 2022.
\]

Thus, concluding

\[
Q < \frac{5}{36}.
\]

Hence, we confirm that

\[
\boxed{Q < \frac{5}{36}}.
\]