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Reference this question: https://ask.sagemath.org/question/35305/how-i-can-test-this-equality-with-sage/

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?

Reference this question: https://ask.sagemath.org/question/35305/how-i-can-test-this-equality-with-sage/

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?

Reference this question: https://ask.sagemath.org/question/35305/how-i-can-test-this-equality-with-sage/

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?

Reference this question: https://ask.sagemath.org/question/35305/how-i-can-test-this-equality-with-sage/

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?

Reference this question: https://ask.sagemath.org/question/35305/how-i-can-test-this-equality-with-sage/

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?