calculer-n-1-oo-artan-2-n-2-

Question Number 181238 by SANOGO last updated on 23/Nov/22

c a l c u l e r

\underset{n = 1}{\sum^{+ o o}} a r t a n (\frac{2}{n^{2}})

Answered by qaz last updated on 23/Nov/22

\arctan \frac{2}{n^{2}} = \arctan (n + 1) - \arctan (n - 1)

Σ \arctan \frac{2}{n^{2}} = Σ a_{n} = Σ a_{2 n - 1} + Σ a_{2 n}

= Σ {[\arctan (2 n) - \arctan (2 n - 2)] + [\arctan (2 n + 1) - \arctan (2 n - 1)]}

= [\arctan (2 \cdot + \infty) - \arctan (2 \cdot 1 - 2)] + [\arctan (2 \cdot + \infty + 1) - \arctan (2 \cdot 1 - 1)]

= \frac{3}{4} π

Commented by mr W last updated on 23/Nov/22

s e e m s t o b e w r o n g . p l e a s e r e c h e c k s i r .

Commented by qaz last updated on 23/Nov/22

0 < \underset{n = 1}{\sum^{\infty}} \arctan \frac{2}{n^{2}} < \underset{n = 1}{\sum^{\infty}} \frac{2}{n^{2}} \approx 3.29

\frac{3}{4} π \approx 2.35, \frac{5}{4} π \approx 3.92

\frac{3}{4} π i s c o r r e c t, i t h i n k

Commented by MJS_new last updated on 23/Nov/22

I also think {it}^{'} s correct

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