Question

Evaluating series. Evaluate the following infinite series or state that the series diverges. sum_{k=1}^inftyfrac{9}{(3k-2)(3k+1)}

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ANSWERED
asked 2020-11-24
Evaluating series. Evaluate the following infinite series or state that the series diverges.
\(\displaystyle{\sum_{{{k}={1}}}^{\infty}}{\frac{{{9}}}{{{\left({3}{k}-{2}\right)}{\left({3}{k}+{1}\right)}}}}\)

Answers (1)

2020-11-25
To find the given sum:
We will split the given fraction into partial fractions first.
Then we substitute the limits from k = 1 to \(\displaystyle\infty\).
Cancel all the possible terms.
Simplify the remaining terms.
Let \(\displaystyle{\frac{{{9}}}{{{\left({3}{k}-{2}\right)}{\left({3}{k}+{1}\right)}}}}={\frac{{{A}}}{{{3}{k}-{2}}}}+{\frac{{{B}}}{{{3}{k}+{1}}}}\) (1)
Multiplying both sides by \(\displaystyle{\left({3}{k}-{2}\right)}{\left({3}{k}+{1}\right)}\)
\(\displaystyle{9}={A}{\left({3}{k}+{1}\right)}+{B}{\left({3}{k}-{2}\right)}\)
Let \(\displaystyle{k}={\frac{{{9}}}{{{3}}}}\Rightarrow{9}={3}{A}\Rightarrow{A}={3}\)
Let \(\displaystyle{k}=-{\frac{{{1}}}{{{3}}}}\Rightarrow{9}=-{3}{B}\Rightarrow{B}=-{3}\)
Substitute these vakues in (1),
\(\displaystyle{\frac{{{9}}}{{{\left({3}{k}-{2}\right)}{\left({3}{k}+{1}\right)}}}}={\frac{{{3}}}{{{3}{k}-{2}}}}-{\frac{{{3}}}{{{3}{k}+{1}}}}={S}{\left[{\frac{{{1}}}{{{3}{k}-{2}}}}-{\frac{{{1}}}{{{3}{k}+{1}}}}\right]}\)
\(\displaystyle{\sum_{{{k}={1}}}^{\infty}}{\frac{{{9}}}{{{\left({3}{k}-{2}\right)}{\left({3}{k}+{1}\right)}}}}={3}{\sum_{{{k}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{3}{k}-{2}}}}-{\frac{{{1}}}{{{3}{k}+{1}}}}\right)}\)
\(\displaystyle={3}{\left[{\left({\frac{{{1}}}{{{3}{\left({1}\right)}-{2}}}}-{\frac{{{1}}}{{{3}{\left({1}\right)}+{1}}}}\right)}+{\left({\frac{{{1}}}{{{3}{\left({2}\right)}-{2}}}}-{\frac{{{1}}}{{{3}{\left({2}\right)}+{1}}}}+\ldots+{\left({\frac{{{1}}}{{\infty}}}-{\frac{{{1}}}{{\infty}}}\right)}\right]}\right.}\)
\(\displaystyle={3}{\left[{\left({1}-{\frac{{{1}}}{{{4}}}}\right)}+{\left({\frac{{{1}}}{{{4}}}}-{\frac{{{1}}}{{{4}}}}\right)}+\ldots{\left({0}-{0}\right)}\right]}\)
\(\displaystyle={3}{\left({1}\right)}\)
\(\displaystyle={3}\)
Therefore, the given infinite series converges and its value is 3.
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