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Sum of the series: SUM=1+1/2+1/3+1/4+1/5....1/n

 
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Sum of the series: SUM=1+1/2+1/3+1/4+1/5....1/n
Kaustubh_Gogoi
6 Mar, 2008 - 06:33 AM
Post #1

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My exams are thr tomorrow plz help me wth ths problem:

SUM=1+1/2+1/3+1/4+1/5....1/n

n should be input by the programmer.
I coded the program as follows:

CODE

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
main()
{
clrscr();
int x=1, y=2, counter;
float sum, number, n;
printf("Enter the value of n: ");
scanf("%f",&n);
for(counter=1;counter<=n;counter++)
{
number=(1/y);
sum=(x+number);
y++;
x=1/(++y);
printf("%f",sum);
}
return 0;
}



Plz help!!!!
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bhandari
RE: Sum Of The Series: SUM=1+1/2+1/3+1/4+1/5....1/n
6 Mar, 2008 - 06:45 AM
Post #2

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0/10 for that. Don't go for exam

Don't mind that

This post has been edited by bhandari: 6 Mar, 2008 - 06:45 AM
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letthecolorsrumble
RE: Sum Of The Series: SUM=1+1/2+1/3+1/4+1/5....1/n
6 Mar, 2008 - 07:22 AM
Post #3

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Well, you will to practice more, if you wish to do good, but here is the code.

cpp

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>


main() {

	int x=1, y=2, n; //n is the integer..see task
	float sum=1.0f;   

	printf("Enter the value of n: ");
	scanf("%d",&n);


	for(;y<=n;) {				
		printf("%f\n",sum); //intermediate sums
		sum = sum+(1.0f/(y++));
	}
	printf("%f\n\n",sum); //final sum
	return 0;
}
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Stepler
RE: Sum Of The Series: SUM=1+1/2+1/3+1/4+1/5....1/n
6 Mar, 2008 - 08:08 AM
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Your code works incorrectly!
Here is how should be:
CODE

#include<stdio.h>
#include<conio.h>
void main(void)
{
clrscr();
int n;
float sum=0;
printf("Enter the value of n: ");
scanf("%d",&n);
for(int i=1;i<=n;i+=1)
sum+=1./i;
printf("%f",sum);
}


I Write programs to order, If there will be questions my ICQ 392172602
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