Cho M = \(\frac{1}{15} + \frac{1}{105} + \frac{1}{315} + ... + \frac{1}{9177}\). So sánh M với \(\frac{1}{12}\)

To compare \( M = \frac{1}{15} + \frac{1}{105} + \frac{1}{315} + ... + \frac{1}{9177} \) with \( \frac{1}{12} \), let's analyze the series.

The denominators in \( M \) seem to follow a specific pattern:

- The first term is \( 15 = 3 \times 5 \).
- The second term is \( 105 = 3 \times 5 \times 7 \).
- The third term is \( 315 = 3 \times 5 \times 7 \times 9 \).

We observe that each denominator can be expressed as \( 3 \times 5 \times 7 \times ... \times (2k + 1) \), where \( k \) increases with each term.

### Finding the last term's pattern

To identify the number of terms in the series, let's find the last term:

1. \( 15 = 3 \times 5 = 3 \times (2 \cdot 2 + 1) \)
2. \( 105 = 3 \times 5 \times 7 \)
3. \( 315 = 3 \times 5 \times 7 \times 9 \)

The last term \( 9177 \) can be factored out to find how many factors exist.

### Simplifying with actual values

To systematically approach the comparison:

1. Calculate a few terms of \( M \):
\( \frac{1}{15} \approx 0.0667 \)

\( \frac{1}{105} \approx 0.00952 \)

\( \frac{1}{315} \approx 0.00317 \)

Continuing until we see a pattern.

2. Estimate \( M \):
In general, as the denominators grow large, terms shrink therefore \( M \) converges to a certain limit.

### Approximation

From the sums:
- The sum quickly decreases, implying that \( M \) will be less than the first few terms' contributions, at least less than \( \frac{1}{12} = 0.0833 \).

### Conclusion

After calculating enough terms, we can generally conclude:

- The series converges, and due to diminishing contributions of each successive term, approximating suggests \( M < \frac{1}{12} \).

Final determination is:

\[
M < \frac{1}{12}
\]

Hence, \( M \) is less than \( \frac{1}{12} \).