# Revision history [back]

### Sage seems to be improperly computing an infinite sum

Here is the evaluation of an infinite sum in sage:

var('n')
f(n) = (-1)^(n+1)/(3*n+6*(-1)^n)
sum(f(2*n)+f(2*n+1),n,0,oo)

1/3*log(2) - 7/9


Evaluating the same sum in Mathematica:

f[n_] := (-1)^(n + 1)/(3*n + 6*(-1)^n)
Sum[f[2*n] + f[2*n + 1], {n, 0, Infinity}]

1/6 (-2 + Log[4])


Sage seems to be giving an incorrect solution. Am I missing something?

### Sage seems to be improperly computing an infinite sum

Here is the evaluation of an infinite sum in sage:

var('n')
f(n) = (-1)^(n+1)/(3*n+6*(-1)^n)
sum(f(2*n)+f(2*n+1),n,0,oo)

1/3*log(2) - 7/9


Evaluating the same sum in Mathematica:

f[n_] := (-1)^(n + 1)/(3*n + 6*(-1)^n)
Sum[f[2*n] + f[2*n + 1], {n, 0, Infinity}]

1/6 (-2 + Log[4])


Sage seems to be giving an incorrect solution. Am I missing something?

 3 retagged slelievre 13636 ●12 ●132 ●269 http://carva.org/samue...

### Sage seems to be improperly computing an infinite sum

Here is the evaluation of an infinite sum in sage:

var('n')
f(n) = (-1)^(n+1)/(3*n+6*(-1)^n)
sum(f(2*n)+f(2*n+1),n,0,oo)

1/3*log(2) - 7/9


Evaluating the same sum in Mathematica:

f[n_] := (-1)^(n + 1)/(3*n + 6*(-1)^n)
Sum[f[2*n] + f[2*n + 1], {n, 0, Infinity}]

1/6 (-2 + Log[4])


Sage seems to be giving an incorrect solution. Am I missing something?

 4 retagged tmonteil 23843 ●26 ●172 ●438 http://wiki.sagemath.o...

### Sage seems to be improperly computing an infinite sum

Here is the evaluation of an infinite sum in sage:

var('n')
f(n) = (-1)^(n+1)/(3*n+6*(-1)^n)
sum(f(2*n)+f(2*n+1),n,0,oo)

1/3*log(2) - 7/9


Evaluating the same sum in Mathematica:

f[n_] := (-1)^(n + 1)/(3*n + 6*(-1)^n)
Sum[f[2*n] + f[2*n + 1], {n, 0, Infinity}]

1/6 (-2 + Log[4])


Sage seems to be giving an incorrect solution. Am I missing something?