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# Tag Archives: sequences

## Sum of reciprocal sequence

Let be a sequence such that and Evaluate the sum Solution

Let us consider the sequence and observe that This in return means, Thus, which is the desired result. Therefore, Notes:

1. The recursive relation has a closed form if-f or . In our case it is: where .

2. The number is known as the Grafting constant.
3. Under the same assumptions it holds that ## Limit of a sequence

Let be a function such that and is differentiable at . Let us set Evaluate the limit .

Solution

Since is differentiable at , there is some such that and is of course continuous.

Thus, Let . There exists such that which in return means that . Hence , for larger than it holds that On the other hand , the sum is a Riemann sum and converges to .

In conclusion, ## Limit of a sum

Let be a real sequence such that , , and . Prove that Solution

Lemma: Let be a bounded sequence of, say, complex numbers and let be another complex number. where the latter means “convergence on a set of density ” i.e. there exists a subsequence such that and is dense i.e. Proof: The proof is omitted because it is too technical.

Hence, All we need is that the logarithm is continuous and that as well as are bounded sequences.

Note: We can simplify the conditions to and .

The exercise along with the solution may be found on AoPS.com . The proof of the claim may also be found there.

## Convergent sequence

Let be a sequence of positive real numbers such that

(1) Prove that converges.

Solution

Fix and let . Then, there exist such that where . Thus, Letting it follows that Since this holds forall it follows that and the result follows.

## On a nested sin sequence

Consider the sequence defined recursively as Prove that .

Solution

Lemma: If is a sequence for which then Proof: In Stolz theorem we set and .

It is easy to see that is is monotonically decreasing to zero. Moreover, an application of L’Hospital’s rule gives Therefore Now, due to the lemma we have and the result follows.

Remark : The asymptotic now follows to be .

Problem: Find what inequality should satisfy such that the series converges.

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