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