With Mathematica 13.1,numerically:
NSum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity},
Method -> "AlternatingSigns", WorkingPrecision -> 32]
(*0.36398547250893341852488170865923*)
NSum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity},
Method -> "WynnEpsilon", NSumTerms -> 10^5, WorkingPrecision -> 32]
(*0.36398547250893341852488170816398*)
Symbolically:
Sum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}]
(*1/2 (1 - \[Pi] Csch[\[Pi]])*)
N[%, 32]
(*0.36398547250893341852488170865923*)
If you use WolframAlfa answer is not precisie ,because for quick answer to you from web.
Try to put:
N[Sum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}] ,32]
Give you:
0.36398547250893341852488170816398