sum of series. Vectorised (no loop)

How can I sum n terms of

1-1/2+1/3-1/4......

Answer 1

[Khalid Mahmood]

% To reduce 2 more lines
function s=vsum(n)
if nargin<1
  s=1; return
end

if mod(n,2)==0, n=n-1;end
s=1+sum(1./[3:2:n] -1./[2:2:n])

[DGM]

This is the same as a prior post. It only gives the correct answer for odd inputs.

[Mohammad Ali]

Khalid's answer is perfect

[DGM]

It's perfect if you want the wrong answer 50% of the time.

This is demonstrable. Just test it.

function s=vsum(n)
    if nargin<1
      s=1; return
    end
    if mod(n,2)==0, n=n-1;end
    s=1+sum(1./[3:2:n] -1./[2:2:n]);
end

Test with odd argument:

s1 = vsum(5)
s2 = 1 - 1/2 + 1/3 - 1/4 + 1/5

results match

s1 =
    0.7833
s2 =
    0.7833

Test with even argument:

s1 = vsum(4)
s2 = 1 - 1/2 + 1/3 - 1/4

results don't match

s1 =
    0.8333
s2 =
    0.5833

This whole thing looks like an attempt to make the vector lengths match when they shouldn't.

s2 = sum(1./(1:2:nt))-sum(1./(2:2:nt))

is simpler and actually correct.

Answer 2

[Mohammad Ali]

I hope you may want this.

function s=vsum(n)
if nargin<1
  s=1; return
end
if mod(n,2)==0, n=n-1;end
nod=[3:2:n]
nev=[2:2:n]
s=1+sum(1./nod-1./nev)

[DGM]

This only gives the correct answer for odd inputs.

Answer 3

[DGM]

You can calculate the sum of a finite alternating harmonic series easy enough:

N = 1000;
n = 1:N;
s = sum((2*mod(n,2)-1)./n)

gives

s =
    0.6926