## An inequality

Let be positive real numbers such that

Prove that

(Vojtech Jarnik / Second Category / 2016)

Solution

Using the AM – GM inequality we have that

as well as . Thus:

## Uniform convergence

Given the sequence of functions where such that

Prove that

(i) the serieses and converge uniformly to functions .

(ii) the functions are continuous.

(iii) .

(iv) it holds that

Solution

(i) This is an immediate consequence of the Weierstrass M Test . We simply note that and of course

(ii) The uniform limit of continuous functions is continuous. This is enough for us to extract that both and throughout .

(iii) For a limit of differentiable functions, a sufficient condition for claiming that the derivative of the limit is the limit of the derivatives is the uniform convergence of the sequence of derivatives. Since the series for converges uniformly, we can claim that .

(iv) For the first integral we have that

and for the second integral we have

since the function is odd.

## On matrices

Let be two matrices with real entries. Prove that

provided all the inverses appearing on the left hand side exist.

(Vojtech Jarnik / 2nd Category/ 2015)

Solution

Let be elements of an arbitrary associative algebra with unit. Then:

## A multiple sum

Evaluate

(Vojtech Jarnik / 2nd Category/ 2011)

Solution

We are applying the classical trick, thus:

## Riemann integrability

Let be defined as:

where is a sequence such that and and be an enumeration of the rationals of the interval . Prove that is Riemann integrable and that .

Solution

We define

Apparently are Riemann integrable and . Also uniformly because coincide except of some points (). Hence , since , we have that:

Thefore is Riemann integrable and thus:

The exercise can also be found in mathematica.gr