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Question: \(a_{n}=3-\sum _{k=1}^{n}\frac{1}{k(k+1)(k+1)!}\), for \(n\in \mathbb{N}\). The limit \(\lim _{n\rig...

an=3−∑k=1n1k(k+1)(k+1)!a_{n}=3-\sum _{k=1}^{n}\frac{1}{k(k+1)(k+1)!}, for n∈Nn\in \mathbb{N}. The limit lim⁡n→∞an\lim _{n\rightarrow \infty }a_{n} is

Answer

The limit cannot be determined with the provided information.

Explanation

Solution

Unfortunately, finding a closed-form expression or a telescoping sum for ∑k=1n1k(k+1)(k+1)!\sum _{k=1}^{n}\frac{1}{k(k+1)(k+1)!} is proving to be exceptionally challenging. Without a simpler form for the sum, determining the limit as nn approaches infinity is not possible with standard techniques.