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Sum 出 bug 的情况其实并不算非常罕见，你这个问题似乎跟SE帖子《Why does Mathematica get the infinite sum of this representation of the taylor series of cos2(x) wrong?》（编号293210）是同类问题，那个问题下面的方法全部适用：
(Sum[x^(4 i + 4 j)/(
E^(2 x) ((4 i)! (4 j)!)), {i, 0, \[Infinity]}, {j, 0, \[Infinity]},
Method -> #1] &) /@ {(*"SequentialFirstToSucceed",
"SequentialBestQuality",*)"ParallelFirstToSucceed",
"ParallelBestQuality", "IteratedSummation"}
Limit[Sum[x^(4 i + 4 j)/(
E^(2 x) ((4 i)! (4 j)!)), {i, l, \[Infinity]}, {j, l, \[Infinity]}],
l -> 0] // Simplify
Normal@Series[
Sum[x^(4 i + 4 j)/(
E^(2 x) ((4 i)! (4 j)!)), {i, l, \[Infinity]}, {j, l, \[Infinity]}],
l -> 0] // Simplify
此bug引入时间是版本7：
(Sum[x^(4 i + 4 j)/(
E^(2 x) ((4 i)! (4 j)!)), {i, 0, \[Infinity]}, {j, 0, \[Infinity]},
Method -> #1] &) /@ {(*"SequentialFirstToSucceed",
"SequentialBestQuality",*)"ParallelFirstToSucceed",
"ParallelBestQuality", "IteratedSummation"}
Limit[Sum[x^(4 i + 4 j)/(
E^(2 x) ((4 i)! (4 j)!)), {i, l, \[Infinity]}, {j, l, \[Infinity]}],
l -> 0] // Simplify
Normal@Series[
Sum[x^(4 i + 4 j)/(
E^(2 x) ((4 i)! (4 j)!)), {i, l, \[Infinity]}, {j, l, \[Infinity]}],
l -> 0] // Simplify
此bug引入时间是版本7：
