## Bernstein limit

Let be real numbers. Consider the function:

as well as the polynomial . Evaluate the limit .

**Solution**

The polynomial is nothing else than a Bernstein one. is Riemann integrable on and is a discontinuity of a first kind. Thus,

where and .

## An iteration limit

Let be a natural number and let where the number of ‘s in the definition of is . For example:

Evaluate the limit:

**Solution**

We easily see by induction that

as well as

Thus,

and thus does not exist.

## Double summation

Let be a real number. Evaluate the series:

**Solution**

Since the summands are all positive , we can sum by triangles. Thus,

where is the Riemann zeta function.

## Inequality of a function

Let be a differentiable function with continuous derivative. Prove that:

**Solution**

For it holds that

Taking absolute values and using basic properties of the integral we get

Integrating we have:

(1)

Working similarly on we get

(2)

Adding equations we get the result.

## An arccotangent integral

Let denote the Riemann zeta function. Prove that

**Solution**

*Background: **I have always been of the opinion that integrals of this form diverged when integrated from something to infinity. Apparently, this is not the case since the above not only converges but it has a nice closed form involving the Riemann zeta function. The technique to break it down , is pretty simple actually. Integration by parts! Yep! We combine IBP along with Fourier series and **voilà. However, somewhere in the middle of this evaluation process we will come across a famous constant , ** , coming literally out of nowhere. But this is what to expect when reducing trigonometric integrals to logarithmic ones.*

We begin our evaluation by stating two lemmata:

**Lemma 1: **It holds that .

*Proof: *Successively, we have that:

where is the Catalan’s constant. Also,

which is the known Fourier series expansion of the function.

**Lemma 2: **It holds that .

*Proof: *Successively, we have that:

Hence,