What is an approximate sum of the sequence

Relevant wiki: Harmonic Number

$\begin{aligned} S & = \frac 12 + \frac 23 + \frac 34 + \frac 45 + \cdots + \frac {99}{100} \\ & = 1 - \frac 12 + 1- \frac 13 + 1- \frac 14 + \cdots + 1 - \frac 1{100} \\ & = 99 - \left(\frac 12 + \frac 13 + \frac 14 + \cdots + \frac 1{100} \right) \\ & = 100 - \left(1 + \frac 12 + \frac 13 + \frac 14 + \cdots + \frac 1{100} \right) \\ & = 100 - {\color{#3D99F6}H_{100}} & \small \color{#3D99F6} H_n \text{ is the }n \text{th harmonic number.} \\ & \approx 100 - \ln 100 - {\color{#3D99F6}\gamma} & \small \color{#3D99F6} \gamma \approx 0.5772 \text{ is the Euler-Mascheroni constant.} \\ & \approx \boxed{94.82} \end{aligned}$

I couldnt do it with Harmonic numbers as i didnt knew it so i wrote a program

int n,i;
float sum=0; 

cout<<"Input n=";
cin>>n;


for (i=1;i<=n;i++)
    sum += (i*1.0)/(i+1);


cout<<"SUM="<<sum;
fflush(stdin);
getchar();
return 0;

OUTPUT=

1
2
3
4

ans = 0
for denom in range(1, 100):
    ans += denom / (denom + 1)
print(ans)

or just:

1

print(sum(denom / (denom + 1) for denom in range(1, 100))) 

The solution can be found by re-writing the sum as 100 - H(100) where H(100) is the 100th Harmonic number

which approximates to 100 - (Gamma + 1/2*100 + ln(100))

H(n) = Gamma + 1/2*n + ln(n) Where Gamma is the Euler-Mascheroni constant .