"How to prove∑n=1∞arctan⁡10n(3n2+2)(9n2−1)=ln⁡3−π4" was answered by our community

Scoopedepalj

Answered question

2022-02-23

How to prove

\sum_{n = 1}^{\infty} \arctan \frac{10 n}{(3 n^{2} + 2) (9 n^{2} - 1)} = \ln 3 - \frac{π}{4}

Alissia Head

Beginner2022-02-24Added 5 answers

Since

\arctan (x) = a r g (1 + i x)

and we can factor

1 + \frac{10 \in}{(3 n^{2} + 2) (9 n^{2} - 1)} = \frac{(1 - \frac{i}{n}) (1 + \frac{i}{3 n - 1}) (1 + \frac{i}{3 n + 1}) (1 + \frac{i}{3 n + 1}) (1 + \frac{i}{3 n})}{1 + \frac{2}{3 n^{2}}}

we have that

\arctan (\frac{10}{(3 n^{2} + 2) (9 n^{2} - 1)})

= \arctan (\frac{1}{3 n - 1}) + \arctan (\frac{1}{3 n}) + \arctan (\frac{1}{3 n + 1}) - \arctan (\frac{1}{n})

Now this becomes a telescoping series:

\sum_{n = 1}^{\infty} \arctan (\frac{10 n}{(3 n^{2} + 2) (9 n^{2} - 1)})

= lim_{m \to \infty} \sum_{n = 1}^{m} [\arctan (\frac{1}{3 n - 1}) + \arctan (\frac{1}{3 n}) + \arctan (\frac{1}{3 n + 1}) - \arctan (\frac{1}{n})]

= - \arctan (1) + lim_{m \to \infty} \sum_{n = m + 1}^{3 m + 1} \arctan (\frac{1}{n})

= - \arctan (1) + lim_{m \to \infty} \sum_{n = m + 1}^{3 m + 1} [\frac{1}{n} + O (\frac{1}{n^{3}})]

= \log (3) - \frac{π}{4}

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