Series Program : [(1^1)/1] + [(2^2)/2] + [(3^3)/3] + [(4^4)/4] + [(5^5)/5] + ... + [(n^n)/n]

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

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

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