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let u_k =1−(1−(1/2^k ))^(n−1) 1)prove that Σ u_k converges 2)let f(x)=1−(1−(1/2^x ))^(n−1) with x≥0 prove that ∀p∈N Σ_(k=1) ^(p+1) u_k ≤∫_0 ^(p+1) f(x)dx≤Σ_(k=0) ^p u_k

$l e t u_{k} = 1 - {(1 - \frac{1}{2^{k}})}^{n - 1}$ $1) p r o v e t h a t Σ u_{k} c o n v e r g e s$ $2) l e t f (x) = 1 - {(1 - \frac{1}{2^{x}})}^{n - 1} w i t h x ⩾ 0$ $p r o v e t h a t \forall p \in N$ $\sum_{k = 1}^{p + 1} u_{k} ⩽ \int_{0}^{p + 1} f (x) d x ⩽ \sum_{k = 0}^{p} u_{k}$

Question Number 40892 Answers: 0 Comments: 0

let B(x,y) =∫_0 ^1 t^(x−1) (1−t)^(y−1) dt withx>0and y>0 prove that B(x,y)= ((Γ(x).Γ(y))/(Γ(x+y)))

$l e t B (x, y) = \int_{0}^{1} t^{x - 1} {(1 - t)}^{y - 1} d t$ $w i t h x > 0 a n d y > 0 p r o v e t h a t$ $B (x, y) = \frac{Γ (x) . Γ (y)}{Γ (x + y)}$

Question Number 40891 Answers: 0 Comments: 1

let x>0 and y>0 and B(x,y) =∫_0 ^1 t^(x−1) (1−t)^(y−1) dt 1)prove that B(x,y)=B(y,x) 2)B(x+1,y)=(x/y) B(x,y+1) 3)B(x+1,y)=(x/(x+y))B(x,y) 4)B(x,n+1)=((n!)/(x(x+1)....(x+n))) 5)B(n,p) = (1/((n+p−1)C_(n+p−2) ^(p−1) ))

$l e t x > 0 a n d y > 0 a n d$ $B (x, y) = \int_{0}^{1} t^{x - 1} {(1 - t)}^{y - 1} d t$ $1) p r o v e t h a t B (x, y) = B (y, x)$ $2) B (x + 1, y) = \frac{x}{y} B (x, y + 1)$ $3) B (x + 1, y) = \frac{x}{x + y} B (x, y)$ $4) B (x, n + 1) = \frac{n!}{x (x + 1) . . . . (x + n)}$ $5) B (n, p) = \frac{1}{(n + p - 1) C_{n + p - 2}^{p - 1}}$

Question Number 40890 Answers: 0 Comments: 2

1)calculate ∫_(1/(n+1)) ^(1/n) [(1/t)−[(1/t)]]dt 2)prove that ∫_0 ^1 [(1/t)−[(1/t)]]dt=1−γ γ is constant number of euler

$1) c a l c u l a t e \int_{\frac{1}{n + 1}}^{\frac{1}{n}} [\frac{1}{t} - [\frac{1}{t}]] d t$ $2) p r o v e t h a t \int_{0}^{1} [\frac{1}{t} - [\frac{1}{t}]] d t = 1 - γ$ $γ i s c o n s t a n t n u m b e r o f e u l e r$

Question Number 40889 Answers: 1 Comments: 0

prove?that ∫_0 ^1 ((1−(1−t)^n )/t)dt =Σ_(k=1) ^n (1/k)

$p r o v e ? t h a t$ $\int_{0}^{1} \frac{1 - {(1 - t)}^{n}}{t} d t = \sum_{k = 1}^{n} \frac{1}{k}$

Question Number 40888 Answers: 0 Comments: 0

prove that ∀ξ ∈]0,π[ ∣∫_ξ ^π (((sint)/t))^n dt∣≤π(((sinξ)/ξ))^n n integr natural

$p r o v e t h a t \forall ξ \in] 0, π [$ $∣ \int_{ξ}^{π} {(\frac{s i n t}{t})}^{n} d t ∣⩽ π {(\frac{s i n ξ}{ξ})}^{n} n i n t e g r n a t u r a l$

Question Number 40887 Answers: 0 Comments: 1

calculate ∫_0 ^1 ((tln(t))/(t^2 −1))dt

$c a l c u l a t e \int_{0}^{1} \frac{t l n (t)}{t^{2} - 1} d t$

Question Number 40886 Answers: 0 Comments: 0

prove that ∫_0 ^1 ((t^(2p+1) ln(t))/(t^2 −1))dt =(π^2 /(24)) −(1/4)Σ_(k=1) ^p (1/k^2 )

$p r o v e t h a t$ $\int_{0}^{1} \frac{t^{2 p + 1} l n (t)}{t^{2} - 1} d t = \frac{π^{2}}{24} - \frac{1}{4} \sum_{k = 1}^{p} \frac{1}{k^{2}}$

Question Number 40885 Answers: 0 Comments: 1

prove that 1) ∫_0 ^1 ((t^p ln(t))/(t−1))dt =(π^2 /6) −Σ_(k=1) ^p (1/k^2 ) 2) ∫_0 ^1 ((t^(2p) ln(t))/(t^2 −1))dt =(π^2 /8) −Σ_(k=0) ^(p−1) (1/((2k+1)^2 ))

$p r o v e t h a t$ $1) \int_{0}^{1} \frac{t^{p} l n (t)}{t - 1} d t = \frac{π^{2}}{6} - \sum_{k = 1}^{p} \frac{1}{k^{2}}$ $2) \int_{0}^{1} \frac{t^{2 p} l n (t)}{t^{2} - 1} d t = \frac{π^{2}}{8} - \sum_{k = 0}^{p - 1} \frac{1}{{(2 k + 1)}^{2}}$

Question Number 40884 Answers: 2 Comments: 0

1) fond ∫_0 ^1 ((ln(t))/(t^2 −1))dt 2) find ∫_0 ^1 ((ln(t))/(t^4 −1))dt

$1) f o n d \int_{0}^{1} \frac{l n (t)}{t^{2} - 1} d t$ $2) f i n d \int_{0}^{1} \frac{l n (t)}{t^{4} - 1} d t$

Question Number 40883 Answers: 1 Comments: 0

find ∫_0 ^∞ (t^p /(e^t −1))dt with p∈N^★

$f i n d \int_{0}^{\infty} \frac{t^{p}}{e^{t} - 1} d t w i t h p \in N^{★}$

Question Number 40882 Answers: 0 Comments: 0

1)prove that ∀n≥2(n inyegr) x^(2n) −1=(x−1)(x+1)Π_(k=1) ^(n−1) (x^2 −2cos(((kπ)/n))x+1) 2)find the value of ∫_0 ^π ln(x^2 −2xcost +1)dt

$1) p r o v e t h a t \forall n ⩾ 2 (n i n y e g r)$ $x^{2 n} - 1 = (x - 1) (x + 1) \prod_{k = 1}^{n - 1} (x^{2} - 2 c o s (\frac{k π}{n}) x + 1)$ $2) f i n d t h e v a l u e o f \int_{0}^{π} l n (x^{2} - 2 x c o s t + 1) d t$

Question Number 40880 Answers: 0 Comments: 0

prove that Σ_(k=n) ^∞ (1/k^α ) ∼ (1/((α−1)n^(α−1) ))with α>1

$p r o v e t h a t \sum_{k = n}^{\infty} \frac{1}{k^{α}} \sim \frac{1}{(α - 1) n^{α - 1}} w i t h α > 1$

Question Number 40878 Answers: 0 Comments: 0

let u_0 >0 and ∀n∈N u_(n+1) =u_n +(1/u_n ) 1) prove that (u_n )is increasing and lim u_n =+∞ 2)by consideringthe functionϕ(t)=(1/(2t+x)) prove that ∀n∈N Σ_(k=1) ^n (1/(2k+x)) ≤(1/2)ln(1+((2n)/x)) 3)find a equivalent of u_n (n→+∞)

$l e t u_{0} > 0 a n d \forall n \in N$ $u_{n + 1} = u_{n} + \frac{1}{u_{n}}$ $1) p r o v e t h a t (u_{n}) i s i n c r e a s i n g a n d l i m u_{n} = + \infty$ $2) b y c o n s i d e r i n g t h e f u n c t i o n φ (t) = \frac{1}{2 t + x}$ $p r o v e t h a t \forall n \in N \sum_{k = 1}^{n} \frac{1}{2 k + x} ⩽ \frac{1}{2} l n (1 + \frac{2 n}{x})$ $3) f i n d a e q u i v a l e n t o f u_{n} (n \to + \infty)$