class Harmonic
{
public static void main(String…s)
{
int n,i;
float sum=0;
n=Integer.parseInt(s[0]);
for(i=1;i<=n;i++)
{
sum=sum+(float)1/i;
}
System.out.println(“nSum=”+sum);
}
}
Java program to find sum of harmonic series 1 + 1/2 + 1/3 + 1/4 + 1/5 +……+ 1/n
class Harmonic
{
public static void main(String…s)
{
int n,i;
float sum=0;
n=Integer.parseInt(s[0]);
for(i=1;i<=n;i++)
{
sum=sum+(float)1/i;
}
System.out.println(“nSum=”+sum);
}
}
Plz tell the code for harmonic series
S = 1-1/2 + 1/3 – 1/4…… -1/10
class Harmonic
{
public static void main(String…s)
{
int n,i;
float sum=0;
n=Integer.parseInt(s[0]);
double sign=1;
for(i=1;i<=n;i++)
{
sum=sum+((float)1/i*sign);
sign*=-1;
}
System.out.println(“nSum=”+sum);
}
}
import java .util.Scanner;;
public class Main
{
public static void main(String[] args) {
Scanner s = new Scanner (System.in);
int n = s.nextInt();
double sum= 0;
for (float i = 1; i<=n;i++){
sum+=(1/i)*Math.pow(-1,i+1);
}
System.out.println(sum);
}
}
I want Harmonic series program for the below expression.
1/(n+1)+1/(n+2)+1/(n+3)+………1/(n+n);
please,help me for this.
still there no update for my request.please send the reply for the request
int s=0;
for(i=1;i<=n;I++)
{
s=s+(1/(n+I));
}
Sopln(s);
How to find the sum of series
Question:
1*1+ 2*2+3*3*3+4*4+5*5+6*6*6+……
public class sumofsquare {
public static void main(String[] args) {
int i,num=10;
int sum=0;
for(i=1;i<=num;i++)
sum += i*i;
System.out.println("SUm of squares : "+sum);
}
}
thanx a lot bro…
why is (float) used in following statement?
sum=sum+(float)1/i;
Because if you divide 1 by i which is an integer your ans will be an integer. E.g. if i=2, 1/i will give you 0 and not 0.5, that’s why you typecast either numerator or denominator to float, so that the division 1/i is also float
import java.util.Scanner;
class sum_of_hp
{
public static void main(String args[])
{
double num,i,sum=0;;
Scanner sc=new Scanner(System.in);
System.out.print(“upto how many terms you want to sum = “);
num=sc.nextInt();
for(i=1;i<=num;i++)
{
sum=sum+(1/i);
}
System.out.println("sum of the series upto "+num+" terms is " +sum);
}
}
instead of using typecasting why didnot you take double type it is easier than type casting
import java.util.Scanner;
class sumofhp
{
public static void main(String args[])
{
double num,i;
double sum=0;
Scanner sc=new Scanner(System.in);
System.out.print(“upto how many terms you want to sum = “);
num=sc.nextInt();
for(i=1;i<=num;i++)
{
sum=sum+(1/i);
}
System.out.println("sum of the series upto "+num+" terms is " +sum);
}
}
instead of taking two different data type int and float can i use double type?
How to find Series of
1+2/3+3/5+4/7….n.
import java.util.Scanner;
public class sumofseries {
public static void main(String[] args)
{
double sum = 0;
int n;
System.out.println(“1!/1+2!/2+3!/3+4!/4+5!/5”);
Scanner s = new Scanner(System.in);
System.out.print(“Enter the no. of terms in series:”);
n = s.nextInt();
sumofseries obj = new sumofseries();
for(int i = 1; i 0)
{
mul = mul * x;
x–;
}
return mul;
}
}
How to find 9th n for the below formula
H(n,r) = 1/1^r + 1/2^r + … + 1/n^r
how to find sum of series 1 +(1+ 1/2^2) +(1+1/ 2^2 +1/3^3 )+ …..+(1+1/ 2^2 +1/3^3 +……+ 1/n^n)
Write a javascript program to create a function which takes an argument n and calculate the sum of the following series: (1+½+⅓+¼+……….+1/n). Use for loop to execute code. Call the function and print output. Your output code should return the output variable.