求最小值

我的cpu在呻吟~~~~

Fold[# + 1/(#2 + Sqrt[#2]) &, 0.0, Range[9881330]] // AbsoluteTiming
如果只迭代，不加条件，速度还行，但是有了条件，就太慢了，怎么破？

FindMinimum[\!\(\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\), \(x\)]\*FractionBox[\(1.`\), \(\*SqrtBox[\(n\)] + n\)]\) - 15, {{x, 10000000}}, Method -> "Newton"]

笨办法一种 @ 草红样

Clear["Global`*"]
time1 = AbsoluteTime[];
g[x_] := NSum[1/(i + Sqrt[i]), {i, x}];
a = 20;
n = 10^a;
judge = 15;
Do[n = 10^(a - j); x = g[n]; If[x < judge, b = a - j; Break[]], {j, a}]
Do[Do[If[g[Sum[f[i] 10^i, {i, b - k, b}] + j 10^(b - k - 1)] < judge &&
g[Sum[f[i] 10^i, {i, b - k, b}] + (j + 1) 10^(b - k - 1)] >=
judge, f = j], {j, 0, 9}], {k, -1, b - 1}]
sol = 1 + Sum[f[i] 10^i, {i, 0, b}]
g[sol]
time2 = AbsoluteTime[];
Print["Time Cost:", time2 - time1]

num = 15;
g[x_] := NSum[1/(i + Sqrt[i]), {i, x}]
b = NestWhile[# + 1 &, 1, g[10^#] < num &] - 1
Do[If[g[Sum[f[i] 10^i, {i, k + 1, b}] + j*10^(k)] < num, f[k] = j],
{k, b, 0, -1}, {j, 0, 9}]
f /@ Range[0, b] // Reverse // FromDigits
简洁了一点，但是还能再优化的，暂时没想到

感觉可以用二分法来做，

a = 9881330;
15 - NIntegrate[1/(
IntegerPart[i + 1] + Sqrt[IntegerPart[i + 1]]), {i, 0, a}]
数值积分，这样的话精确度不够

g[n_] := NSum[1/(i + Sqrt[i]), {i, n}] - 15;a = 1;b = 10^10;Do[If[g[IntegerPart[(a + b)/2]] > 0, b = IntegerPart[(a + b)/2], a = IntegerPart[(a + b)/2]]; If[ b-a==1,Print[ b];Break[]];,{i,100}]
@ 孺子剑牛不群
二分法，谁整理一下，话说度娘怎么左中括号和字母b连在一起就会吞掉呢？？

二分法搞出来了
t1 = AbsoluteTime[];
num = 15;
f[x_] := NSum[1/(i + Sqrt[i]), {i, x}];
n = NestWhile[# + 1 &, 1, f[10^#] < num &]
min = 10^(n - 1);
max = 10^n;
mid = Round[(min + max)/2];
While[Abs[f[mid] - num] > 10^-6, If[f[mid] >= num, max = mid - 1];
If[f[mid] <= num, min = mid + 1];
mid = Floor[(min + max)/2];];
{min, mid, max}
AbsoluteTime[] - t1
Clear["`*"]