Question

recall that for an infinite series \\(\sum_{i = 1}^{\infty} a_i\\), the \\(k\\)th partial sum is given by \\(s_k = \sum_{i = 1}^{k} a_i\\). a series converges if the sequence of partial sums converge, that is, \\(s_k \to s\\) as \\(k \to \infty\\).

to determine if \\(\sum_{n = 1}^{\infty} \frac{1}{(n + 4)(n + 5)}\\) converges, we first find an explicit formula for \\(s_k\\).

following the hint, we use partial fraction decomposition to rewrite the terms of the series.

\\\frac{1}{(n + 4)(n + 5)} = \frac{a}{n + 4} + \frac{b}{n + 5}\\
\\= \frac{(a + b)n + (5a + 4b)}{(n + 4)(n + 5)}\\

equating coefficients in the numerators, we obtain the following system.

\\(a + b =\\)

\\(5a + 4b =\\)

solving this system, we have \\(a =\\) , \\(b =\\) . therefore, we can rewrite the series as follows.

\\(\sum_{n = 1}^{\infty} \frac{1}{(n + 4)(n + 5)} = \sum_{n = 1}^{\infty} \frac{\text{ }}{n + 4} + \sum_{n = 1}^{\infty} \frac{\text{ }}{n + 5}\\)

Explanation:

Set up the system of equations

$$
LATEXBLOCK0
$$

Solve for A and B

$$
LATEXBLOCK1
$$

Rewrite the series terms

$$ \sum_{n=1}^{\infty} \frac{1}{(n+4)(n+5)} = \sum_{n=1}^{\infty} \frac{1}{n+4} + \sum_{n=1}^{\infty} \frac{-1}{n+5} $$

Answer:

Equating coefficients in the numerators, we obtain the following system.
\(A + B =\) <blank>0</blank>
\(5A + 4B =\) <blank>1</blank>

Solving this system, we have \(A =\) <blank>1</blank>, \(B =\) <blank>-1</blank>. Therefore, we can rewrite the series as follows.

$$\sum_{n=1}^{\infty} \frac{1}{(n+4)(n+5)} = \sum_{n=1}^{\infty} \frac{\text{1}}{n+4} + \sum_{n=1}^{\infty} \frac{\text{-1}}{n+5}$$