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]*/ 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; }