Sum series using function and loop (C++) - c++

Question

I am writing some code that prints out a sum series using a loop and a function.
I intend the equation to look like this
m(i) = (1/2) + (2/3) + ... (i / i + 1)
The problem is that my code always gives me incorrect answers and not printing what it's supposed to. For example, when I input 1 into 1 the answer should be 0.5
This is my code:
#include <iostream>
using namespace std;
void sumSeries(int x);
int main() {
sumSeries(1);
return 0;
}
void sumSeries(int x){
double sum = 0;
for(int i = 0; i < x; i++){
sum = (x/x + 1);
sum += sum;
}
cout<<sum;
}

Answer 1

Indeed, you overwrite your sum but also take care of your integer division.
You may change it as sum += i/(double)(i + 1);
#include <iostream>
using namespace std;
void sumSeries(int x);
int main() {
sumSeries(5);
return 0;
}
void sumSeries(int x){
if (x<0)
{
return;
}
double sum = 0;
for(int i = 0; i < x; i++){
sum += i/(double)(i + 1);
}
cout<<sum;
}

Answer 2

I see two problems in your code.
First: (x/x+1) != (x/(x+1)), in this case C++ obeys the normal point before line calculation rules.
Second: You are overwriting your sum in each iteration, instead of that you should direct add to sum: sum+=x/(x+1)
And a third issue, as noted by Simon Kraemer, is that you are using integer division, to get the correct results you must cast at least one of the operands to a floating point number.

Answer 3

What you want is:
void sumSeries(int x){
double sum = 0;
for(int i = 1; i <= x; i++){ // include i in the list
sum += static_cast<double>(i)/(i + 1); // force the operation as double
}
cout<<sum;
}

Answer 4

your mathematical expression has something not normal. Do you mean M(i)= sum(1-i){i/i+1}? , or 1/2 and 1/3 are constants?
in your case as gerum answered it is a small Operator Precedence problem to learn how the C++ compiler prioritize the operators follow here.
your function also should have a guard against zero denominator (undefined values).
Also you should observe that you take int/int division which will ignore the remaining value. then you should consider that by converting the numerator or the denominator to double before the division here .
then your code should be:
#include <iostream>
using namespace std;
void sumSeries(int x);
int main() {
sumSeries(1);
return 0;
}
void sumSeries(int x){
double sum = 0;
for(int i = 0; i < x; i++){
if ((x+1)!=0){
sum += (double)x/(x + 1);
}
// the else will apply only if x==-1
else {
cout<<"the denominator is zero"<<endl;
throw;
}
}
cout<<sum;
}