Find the sum of the first n

Find the sum of the first n terms of the harmonic series where n is an integer greater than one 1+1/2+1/3+1/4+1/5+.....

Answer 1

[John Chilleri]

Hello,

This can be done simply with,

n = 100; % whatever you want
sum_harm = 0;
for i = 1:n
   sum_harm = sum_harm + 1/i;
end

or even,

n = 100; % whatever you want
sum_harm = sum(1./(1:n));

Hope this helps!

[bassam792]

it's wrong

[John Chilleri]

Did you change the n to your desired value?

[John Chilleri]

There is a built-in harmonic function, but I wasn't sure if you wanted to write your own,

harmonic(n)

is the built-in MATLAB function, which should agree with both above.

Hope this helps!

[John Chilleri]

Fixed my code - I was using a variable named sum, which can cause errors with an unclear workspace, I apologize.

[Marvin Jones]

What if you wanted to go in reverse say from n to 1?

[Walter Roberson]

The theoretical sum would be the same. If you are using floating point then the result could differ. The order of operations of built-in functions like harmonic() is not specified. For doing it yourself, replace 1:n by n:-1:1

[Walter Roberson]

format long g
n = 1e10; % whatever you want
sum_harm_forward = 0;
sum_harm_reverse = 0;
for i = 1:n
   sum_harm_forward = sum_harm_forward + 1/i;
   sum_harm_reverse = sum_harm_reverse + 1/(n - i + 1);
end

sum_harm_forward

sum_harm_forward = 
          23.6030665949975

sum_harm_reverse

sum_harm_reverse = 
          23.6030665948883

sum_harm_forward - sum_harm_reverse

ans = 
        1.092317347684e-10

Answer 2

[Roger Stafford]

Or how about this one-liner:

H = det(diag(2:n)+ones(n-1))/factorial(n);

inspire by http://oeis.org/A000254.

[Ahmad Taheri]

hi ,admin

one qustions

∑1/n=?

∑1/n^2=??

[Walter Roberson]

∑1/n is infinite.

∑1/n^2 is pi^2/6

Answer 3

[Walter Roberson]

syms x n
symsum(1/x, x, 1, n)

[Hesbon Osoro]

Very much useful to test the convergence of a harmonic series without lagging a machine, so fast code also. It is of great help.