What is the Limit of the Sequence {arctan(2n)} and Does it Converge to pi/2?

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The sequence {arctan(2n)} is bounded by π/2 and converges to π/2 due to its monotonic increasing nature. The limit of the function f(x) = L confirms that the sequence approaches π/2 as n approaches infinity. However, it is essential to note that being bounded alone does not guarantee convergence, as illustrated by the sequence 1, 1/2, 1/4, 1/8, which is bounded by 1 but does not converge to it. The convergence theorem states that the sequence converges to the supremum of {arctan(2n): n>0}, which is indeed π/2.

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For the sequence {arctan2n}, it is bounded by pi/2, so does that mean it converges to pi/2, because the limit of f(x) = L, and therefore the limit of the sequence is pi/2?
 
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Here is an example of a sequence: 1, 1/2, 1/4, 1/8, ...

It is bounded by 1, but does not converge to 1.

Therefore, you need other conditions.
 
Your other condition is that its monotonic increasing in n. And there's a theorem that says it converges to the sup of {arctan(2n): n>0}, which is pi/2.
 

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