>>>971
>sin(π√k)をマクローリン展開してから、和の順序を変えて煤緻/√n、(√k)^3)/√n,,,
sin(π√k)=π√k-(π√k)^3/3!+(π√k)^5/5!-(π√k)^7/7!+…
Σ[k=1,n]sin(π√k)/√n=πΣ√k/√n-π^3/3!Σ√k^3/√n+π^5/5!Σ√k^5/√n-π^7/7!Σ√k^7/√n+…
=πnΣ√(k/n)/n-π^3n^2/3!Σ√(k/n)^3/n+π^5n^3/5!Σ√(k/n)^5/n-π^7n^4/7!Σ√(k/n)^7/n+…
>をそれぞれn∫[0->1]x^(1/2)dx, n^2∫[0->1]x^(3/2)dx、、、、と積分で置き換えれば、
Σ[k=1,n]√(k/n)^m/n=Σ[k=1,n](k/n)^(m/2)/n≒∫[0,1]x^(m/2)dx=2/(m+2)
Σ[k=1,n]sin(π√k)/√n≒(2/3)πn-(2/5・3!)π^3n^2+(2/7・5!)π^5n^3-(2/9・7!)π^7n^4+…
=(2/π){(1/3)(π√n)^2-(1/5・3!)(π√n)^4+(1/7・5!)(π√n)^6-(1/9・7!)(π√n)^8+…}
=(2/π^2√n){(1/3)(π√n)^3-(1/5・3!)(π√n)^5+(1/7・5!)(π√n)^7-(1/9・7!)(π√n)^9+…}
{(1/3)x^3-(1/5・3!)x^5+(1/7・5!)x^7-(1/9・7!)x^9+…}'=x^2-x^4/3!+x^6/5!-x^8/7!+…
=x(x-x^3/3!+x^5/5!-x^7/7!+…}
=xsin x
(1/3)x^3-(1/5・3!)x^5+(1/7・5!)x^7-(1/9・7!)x^9+…=∫[0,x]xsinx dx
=sinx-xcosx
Σ[k=1,n]sin(π√k)/√n≒(2/π^2√n){sin(π√n)-(π√n)cos(π√n)}
=(2/π){sinc(π√n)-cos(π√n)}
>n→∞で S_n → -(2/π)cos(π√n) となる。
(2/π){1-cos(π√n)}