Answer :

Final answer:

The given conditions relate to the concept of limits in sequences or series in Mathematics. These conditions might be part of a larger mathematical problem or proof, such as determining series convergence. However, more context is needed to provide a thorough answer.

Explanation:

The question you're asking seems to relate to the concept of limits in Mathematics, specifically within the domain of sequences and series. However, the information provided is a bit disjointed and some parts are unclear.

Generally speaking, if we are given that the limit as n approaches infinity of y1 is 0 and yn greater than or equal to n for all n within the Natural numbers (N), these statements could be part of a larger mathematical problem or proof. For example, these conditions might arise when studying series convergence/divergence or in the definition of a particular type of sequence.

But without more context or specific question, it's hard to provide a detailed or concrete answer. Could you please provide more information or a specific problem where these conditions are applied?

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Rewritten by : Barada

The proof is that the limit of any sequence `ynna` is equal to zero. This is based on the fact that the given sequence `y₁` has the limit of zero, and the difference between `ynna` and zero becomes smaller and smaller as n becomes larger and larger.

The given question is about the limit of a sequence. It is a two-part question. The first part is stating that the limit of a sequence `y₁ = 0`. The second part is asking to show that the limit of any sequence `ynna` for all `n ∈ N`, is equal to 0. Let's discuss each part of the question in detail. What is the limit of sequence y₁?The first part of the question is quite simple. The given sequence `y₁` has only one term, and that is zero. Therefore, the limit of sequence `y₁` is also zero. How to show that lim(n→∞) ynna = 0?To show that the limit of sequence `ynna` is equal to zero, we have to prove that the difference between `ynna` and zero becomes smaller and smaller as n becomes larger and larger. That means, for any given ε > 0, we have to find a natural number N such that |ynna - 0| < ε for all n > N. is the proof: Given ε > 0For ε > 0, we need to find a natural number N such that |ynna - 0| < ε for all n > N. The sequence `ynna` is given to be arbitrary, which means that we don't know the exact formula for `ynna`. But we know that lim `ynna` = 0. Therefore, for any given ε > 0, we can find a natural number N such that |ynna - 0| < ε/2 for all n > N. Now, let's use the triangle inequality and conclude the proof:|ynna - 0| = |ynna - y₁ + y₁ - 0| ≤ |ynna - y₁| + |y₁ - 0| < ε/2 + ε/2 = εTherefore, by definition, lim(n→∞) ynna = 0.

The proof is that the limit of any sequence `ynna` is equal to zero. This is based on the fact that the given sequence `y₁` has the limit of zero, and the difference between `ynna` and zero becomes smaller and smaller as n becomes larger and larger.

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