Question

ゲスト

次の極限値を求めよ
お願いします

Answer 1

坂系⊿

(1)lim(Σk)^5/(Σk^4)^2=25/32
(2)log{1/n{(4n)!/(3n)!}^{1/n)}
=-logn+{log(4n)!/(3n)!}/n
=-logn+{log(4n)+log(4n-1)+…log(3n+1)}/n
={-nlogn+log(4n)+log(4n-1)+…log(3n+1)}/n
={log(4n)-logn+log(4n-1)-logn+…log(3n+1)-logn}/n
={log(4)+log((4n-1)/n)+…log((3n+1)/n)}/n
=∫{3→4}logxdx
=[xlogx-x][3→4]=4log4-3log3-4+3=log(256/27)-1
=log(256/27e)
(3)π/n・[1/{sin(π(n+1)/4n)}+1/sin(π(n+2)/4n)}+……+1/{sin(π(n+n)/4n)}
＝π/n・[1/{sin(π(1/4+1/4n)}+1/sin(π(1/4+2/4n)}+……+1/{sin(π(1/4+n/4n)}
＝π/n・[Σ[k=1,n]1/{sin{(π/4)・(1+k/n)}]

lim[n→∞]π/n・[1/{sin(π(n+1)/4n)}+1/sin(π(n+2)/4n)}+……+1/{sin(π(n+n)/4n)}]
＝π lim[n→∞]・(1/n)[Σ[k=1,n]1/{sin{(π/4)・(1+k/n)}]
＝π∫[0,1]dx/sin{π(1+x)/4}
θ＝π(1+x)/4に置換
dθ＝(π/4)dxより、
π∫[π/4,π/2](4/π)dθ/sinθ
＝4∫[π/4,π/2]dθ/sinθ
＝4∫[π/4,π/2]sinθdθ/(sinθ)^2
＝4∫[π/4,π/2]sinθdθ/{1–(cosθ)^2}

t＝cosθに置換
dt＝–sinθdθ

4∫[1/√2, 0]–dt/{1–t^2}
＝4∫[0, 1/√2]dt/(1+t)(1−t)
＝2∫[0, 1/√2]{1/(1−t)–1/(1+t)}dt
＝2[−log(1−t)−log(1+t)][t=0,1/√2]
＝２log[1/{(1−1/√2)・(1+1/√2)}]
＝２log2

2・log2

ゲスト

ありがとうございます。
（1）の式変形を教えて頂けると助かります

坂系⊿

(1)区分求積法から
{(1+2+3+...+n)^5}/(n^10)
= {(1/n)Σ[k=1,n] (k/n) }^5
→ {∫[0,1] x dx}^5 [n→∞]
= 1/32
{(1^4+2^4+3^4+...+n^4)^2}/(n^10)
= {(1/n)Σ[k=1,n] (k/n)^4 }^2
→ {∫[0,1] x^4 dx}^2 [n→∞]
= 1/25
(1/32)/(1/25)＝25/32

ゲスト

ありがとうございます