## The limit is zero

Let be an integrable and uniformly continuous function. Prove that.

Solution

This exercise is quite known. It first made its appearance as a Berkley exam question.

Suppose that the limit is not zero. Then let us pick a sequence and an such that foreach to both hold and . Since is uniformly continuous there exists such that

However in the interval it holds that

foreach positive. The last is a consequence of . And of course this contradicts the fact that the function is integrable. Hence the conclusion.

Note: We can’t drop the condition of uniform continuity. If so, then the function is a counterexample. Indeed it is continuous, integrable but the limit at infinity is not .

## An integral inequality

Let be a continuous function such that

(1)

Prove that .

Solution

We note that satisfies all conditions. Thus:

and thus the conclusion.

## On a determinant

Let be a prime number and let be  a primitive p-th root of unity. Define:

Evaluate the rational number .

Solution

The -th entry of is  . Thus the -th entry of is equal to:

since it is known that for any -th root of unity rather than . Thus:

Thus is there is a at the upper left corner and ‘s along the anti diagonal in the lower right block. Thus:

## A limit of a sum

Evaluate the limit

Solution

This is a very well known limit and someone could argue that the value of it is . Unfortunately, this is not the case. We shall see that the limit equals .

The fastest way is by making use of probabilities. Let us consider independent Poisson distributions with parameter . Then is Poisson with parameter . From the central limit theorem converges in distribution to the standard normal distribution. In particular, if follows a standard normal distribution then

and the conclusion follows.

## An infinite product with Fibonacci

Compute the product

where is the -th Fibonacci sequence term.

Solution

We recall the Cassini identity that the Fibonacci sequence obeys to, hence

where is the golden ratio and its value is .