Series Program 8

[(1^1)/1!] + [(2^2)/2!] + [(3^3)/3!] + [(4^4)/4!] + [(5^5)/5!] + ... + [(n^n)/n!]

double power(int a, int b)
{
    long i, p=1;
    for(i=1;i<=b;i++)
    {
        p=p*a;
    }
    return p;
}

double fact(int n)
{
    long i, f=1;
    for(i=1;i<=n;i++)
    {
        f=f*i;
    }
    return f;
}

int main()
{
    long i,n;
    double sum=0;
    n=5;
    for(i=1;i<=n;i++)
    {
        sum=sum+power(i,i)/fact(i);
    }
    printf("Sum: %lf",sum);
    return 0;
}

Related Links:
- More Series Programs
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