Find out the answer to "Evaluating series. Evaluate the following infinite series or..."

SchachtN

Answered question

2020-11-24

Evaluating series. Evaluate the following infinite series or state that the series diverges.

\sum_{k = 1}^{\infty} \frac{9}{(3 k - 2) (3 k + 1)}

2abehn

Skilled2020-11-25Added 88 answers

To find the given sum:
We will split the given fraction into partial fractions first.
Then we substitute the limits from k = 1 to

\infty

.
Cancel all the possible terms.
Simplify the remaining terms.
Let

\frac{9}{(3 k - 2) (3 k + 1)} = \frac{A}{3 k - 2} + \frac{B}{3 k + 1}

(1)
Multiplying both sides by

(3 k - 2) (3 k + 1)

9 = A (3 k + 1) + B (3 k - 2)

Let

k = \frac{9}{3} \Rightarrow 9 = 3 A \Rightarrow A = 3

Let

k = - \frac{1}{3} \Rightarrow 9 = - 3 B \Rightarrow B = - 3

Substitute these vakues in (1),

\frac{9}{(3 k - 2) (3 k + 1)} = \frac{3}{3 k - 2} - \frac{3}{3 k + 1} = S [\frac{1}{3 k - 2} - \frac{1}{3 k + 1}]

\sum_{k = 1}^{\infty} \frac{9}{(3 k - 2) (3 k + 1)} = 3 \sum_{k = 1}^{\infty} (\frac{1}{3 k - 2} - \frac{1}{3 k + 1})

= 3 [(\frac{1}{3 (1) - 2} - \frac{1}{3 (1) + 1}) + (\frac{1}{3 (2) - 2} - \frac{1}{3 (2) + 1} + \dots + (\frac{1}{\infty} - \frac{1}{\infty})]

= 3 [(1 - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{4}) + \dots (0 - 0)]

= 3 (1)

= 3

Therefore, the given infinite series converges and its value is 3.

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