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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:**

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

- The number is known as the
**Grafting constant**. - 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.