Question

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

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

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.