## Is f necessarily constant?

Let be a differentiable function such that forall . Does it follow that is necessarily constant?

Solution

Since forall it follows that forall . Let and suppose that . Since is dense there will exist a sequence of rational numbers that it converges to . Thus:

contradicting what we had assumed in the first place. Hence is constant.

## A contour integral

Let such that . Prove that:

Solution

We have successively:

## Convergence of series

Examine if the series:

converges.

Solution

We are using the basic inequality

as well as the Hermite Hadamard inequality. Applying the Hermite – Hadamard inequality we have that

Exponentiating we get that

and hence

This leads us to the conclusion that the series diverges.

## Constant sign

Let be a non zero continuous function on a connected surface . Prove that does not change sign on .

Solution

Suppose that and for some . Obviously because if then

and this is an obscurity. Since is connected there exists a continuous curve such that

Define . This function is obviously continuous. Note that . So by Bolzano’s theorem, there exists a such that . This, in return means that but this is impossible since has a constant sign.

## Double series

Evaluate the series:

(Putnam Competition , 2016)

Solution

Since the double series the series converges absolutely we can interchange the summation. Thus: