The sum of the following series | User given limit | 1-2^2/2!+3^3/3!- . . . . . . . . | C-Program

 The sum of the following series: 1 - 2^2 / 2! + 3^3 / 3! - . . . . . . .

Code:- 

#include<stdio.h>
#include<math.h>    // for pow( );
int fact(int a)
{ int ans=1;
  for(int i=1;i<=a;i++)
  { ans *= i; }
  return ans;
}
void main()
{ int i,x;
  float sum=1;
  int fact(int a);
  printf("Enter the number :- ");
  scanf("%d",&x);
  printf("\n1");
  for(i=2;i<=x;i++)
  {
    if(i%2==0)
    { printf(" - %d^%d/%d!",i,i,i);}
    else
    { printf(" + %d^%d/%d!",i,i,i); }
  }
  for(int j=2;j<=x;j++)
  { if(j%2==0)
    { sum-=pow(j,j)/fact(j);}
    else
    { sum+=pow(j,j)/fact(j); }
  }
  printf("  = %f\n",sum);
} 

 

Output:

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