We have an answer to "Find the value of∑n=0∞19n2+9n+2"

Kris Patton

Answered question

2022-02-25

Find the value of

\sum_{n = 0}^{\infty} \frac{1}{9 n^{2} + 9 n + 2}

faraidz3i

Beginner2022-02-26Added 10 answers

Note that

\sum_{n = 0}^{\infty} \frac{1}{9 n^{2} + 9 n + 2} x^{3 n} = \sum_{n = 0}^{\infty} [\frac{1}{3 n + 1} x^{3 n} - \frac{1}{3 n + 2} x^{3 n}]

Define

f (x) = \sum_{n = 0}^{\infty} \frac{1}{3 n + 1} x^{3 n}, g (x) = \sum_{n = 0}^{\infty} \frac{1}{3 n + 2} x^{3 n}

It is easy check that

{(x f (x))}^{'} = \frac{1}{1 - x^{3}}, {(x^{2} g (x))}^{'} = \frac{x}{1 - x^{3}}

lim_{x \to 1} f (x) - g (x) = lim_{x \to 1} \frac{1}{x} \int_{0}^{X} \frac{1}{1 - t^{3}} d t - \frac{1}{x^{2}} \int_{0}^{x} \frac{t}{1 - t^{3}} d t

= lim_{x \to 1} \int_{0}^{1} \frac{1 - u}{1 - {(u x)}^{3}} d u

= \int_{0}^{1} \frac{1 - u}{1 - u^{3}} d u

\int_{0}^{1} \frac{1}{1 + u + u^{2}} d u

= \frac{π}{3 \sqrt{3}}

Tate Puckett

Beginner2022-02-27Added 6 answers

You can use the residue theorem. Rewrite the sum as

\frac{1}{18} \sum_{n = - \infty}^{\infty} \frac{1}{(n + \frac{1}{3}) (n + \frac{2}{3})}

Then use the fact that a consequence of the residue theorem is

\sum_{n = - \infty}^{\infty} \frac{1}{(n + \frac{1}{3}) (n + \frac{2}{3})} = - π \sum_{k} R e s_{z = z_{k}} \frac{\cot π z}{(z + \frac{1}{3}) (z + \frac{2}{3})}

= π [\frac{\cot (\frac{π}{3})}{\frac{1}{3}} - \frac{\cot (2 \frac{π}{3})}{\frac{1}{3}}]

= 2 \sqrt{3} π

Therefore the sum is

\sum_{n = 0}^{\infty} \frac{1}{9 n^{2} + 9 n + 2} = \frac{π}{3 \sqrt{3}}

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