Advanced-Calculus-simplify-n-k-1-2n-1-log-1-tan-kpi-4-2n-1-moreover-find-the-value-of-lim-n-n-n-

Question Number 137397 by mnjuly1970 last updated on 02/Apr/21

\dots \dots . A d v a n c e d \dots \dots \dots C a l c u l u s \dots \dots .

s i m p l i f y :::

Ω_{n} = \underset{k = 1}{\sum^{2 n + 1}} l o g (1 + t a n (\frac{k π}{4 (2 n + 1)}))

m o r e o v e r, f i n d t h e v a l u e o f ::

Ω = {l i m}_{n \to \infty} \frac{Ω_{n}}{n} = ? ? ?

Answered by mindispower last updated on 02/Apr/21

= \underset{k = 0}{\sum^{2 n + 1}} l o g (1 + t g (\frac{k π}{4 (2 n + 1)}) = \underset{k = 0}{\sum^{2 n + 1}} l o g (1 + t g (\frac{π}{4} - \frac{k π}{4 (2 n + 1)}))

∴ l n (1) = 0, \underset{l = m}{\sum^{p}} f (l) = Σ f (p + m - l)

= \underset{k = 0}{\sum^{2 n + 1}} l o g (1 + \frac{1 - t g (\frac{k π}{4 (2 n + 1)})}{1 + t g (\frac{k π}{4 (2 n + 1)})})

= \underset{k = 0}{\sum^{2 n + 1}} l o g (\frac{2}{1 + t g (\frac{k π}{4 (2 n + 1)})}) = \underset{k = 0}{\sum^{2 n + 1}} l o g (2) -^{} Ω_{n}

\Rightarrow 2 Ω_{n} = (2 n + 2) l o g (2) \Rightarrow Ω_{n} = (n + 1) l o g (2)

\frac{Ω_{n}}{n} = (1 + \frac{1}{n}) l o g (2) \Rightarrow \lim_{n \to \infty} \frac{Ω_{n}}{n} = l o g (2)

Commented by mnjuly1970 last updated on 02/Apr/21

v e r y n i c e m r p o w e r

g r a t e f u l \dots

Answered by mnjuly1970 last updated on 02/Apr/21

Ω_{n} = \underset{k = 0}{\sum^{n}} l o g (1 + t a n (\frac{k π}{4 (2 n + 1)})) + \underset{k = n + 1}{\sum^{2 n + 1}} l o g (1 + t a n (\frac{k π}{4 (2 n + 1)}))

= \underset{k = 0}{\sum^{n}} l o g (1 + t a n (\frac{k π}{4 (2 n + 1)})) + \underset{k = 0}{\sum^{n}} l o g (1 + t a n (\frac{(2 n + 1 - k) π}{4 (2 n + 1)}))

= \underset{k = 0}{\sum^{n}} l o g (1 + t a n (\frac{k π}{4 (2 n + 1)})) + \underset{k = 0}{\sum^{n}} l o g (1 + \frac{1 - t a n (\frac{k π}{4 (2 n + 1})}{1 + t a n (\frac{k π}{4 (2 n + 1)})})

= (n + 1) l o g (2) \dots . ✓

\dots \frac{Ω_{n}}{n} = \frac{n + 1}{n} l o g (2) \Rightarrow {l i m}_{n \to \infty} \frac{Ω_{n}}{n} = l o g (2)

\dots . . Ω = l o g (2) \dots . . ✓

h i n t : \underset{r = k}{\sum^{n}} a (r) = a (k) + a (k + 1) + \dots + a (n)

= \underset{r = 0}{\sum^{n - k}} a (n - r) ✓