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

Below is C program for Series Program- [(1^1)/1!] + [(2^2)/2!] + [(3^3)/3!] + [(4^4)/4!] + [(5^5)/5!] + ... + [(n^n)/n!]

Program:

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/* Series Program-[(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; }