False. A sequence "\\{a_n\\}_{n\\in \\mathbb{N}}" is called strictly monotonically decreasing if and only if "a_{n+1} < a_n" for all "n\\in \\mathbb{N}". Since "-(n+1) = -n-1 < -n", the sequence "\\{-n\\}_{n\\in \\mathbb{N}}" satisfies this definition and is strictly monotonically decreasing. Assume that it converges to some "a\\in \\mathbb{R}". Then there is "n_0\\in \\mathbb{N}" such that "a-1 < -m < a+1" for all "m\\geq n_0". By the Archimedean property of the real numbers, there is "n_1\\in \\mathbb{N}" such that "-n_1 \\leq a-1". Letting "n_2 = \\operatorname{max}(n_0, n_1)", we have that "a-1 < -n_2" and "-n_2 \\leq a-1", a contradiction. Therefore, "\\{-n\\}_{n\\in \\mathbb{N}}" is not convergent.
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