Question

Let \( u_n = \frac{18n + 3}{(3n - 1)^2 (3n + 2)^2} \), \( n \in \mathbb{N} \). Then \[ \sum_{n=1}^{\infty} u_n \quad \text{equals} \]

Hint:

For series with complex denominators, consider simplifying or recognizing patterns that may allow the use of known series sum formulas or telescoping series techniques.

Answer 1

Correct Answer: 0.25

Step 1: Analyzing the series.
We are given the series \( u_n \) defined by: \[ u_n = \frac{18n + 3}{(3n - 1)^2 (3n + 2)^2}. \] Our goal is to find the sum of this infinite series.
Step 2: Simplifying the terms.
We begin by simplifying the expression for \( u_n \). Observe that the terms in the numerator and denominator suggest the possibility of a telescoping series. To proceed, let us factor the denominator to analyze the pattern more clearly.
Step 3: Checking the sum of the series.
The series can be recognized as a standard series with a known sum. By applying techniques for summing such series (for example, using partial fractions or recognizing a known pattern), we find that the sum of the series converges to \( \frac{1}{4} \), which is approximately 0.25.
Step 4: Conclusion.
Thus, the sum of the infinite series is approximately \( 0.25 \).