## SEEMOUS 2016 Problem 4 – Solution

**Problem 4.** Let be an integer and set

Prove that

a)

b) .

*Proof:* First, let’s note that each of the integrals is convergent since . We can deduce a better upper bound using the fact that . We have

Now, note that the series of functions

converges uniformly on each interval of the form , since we have a geometric series multiplied by . Indeed, if we denote

then this series is absolutely convergent for and is derivable term by term. Thus, we have . Integrating and noting that we obtain for . Thus, we can invert the series with the integrals and obtain that

which gives

Since we conclude that the series is convergent and thus we can pass to the limit as in the above equality of integral to deduce that

This proves that b) follows once we have a).

To prove a) we make the change of variables and we deduce that

We note that for and that by applying a similar method as in the first part we can switch the series and the integral to conclude that

Noting that we perform an integration by parts in the above integral and we obtain

To finish it is enough to prove that

We have

which by the change of variables gives

Since and the series converges uniformly for we have

since . This finishes the proof.

Alternative proof for a): $I_{n+1}=int_0^{pi/2} y cos^{2n}(y) dy=int_0^{pi/2} y cos^{2n-1}(y) (sin(y))’ dy$ and integrating by parts we get: $I_{n+1}= – 1/(2n)+(2n-1)I_n – (2n-1)I_{n+1}$ which is equivalent with: $I_n/n = 2(I_n – I_{n+1})-1/(2n^2)$. Now, the series is equal to 2I_1-pi^2/12=pi^2/6.

Alternative proof for a): $I_{n+1}=int_0^{pi/2} y cos^{2n}(y) dy=int_0^{pi/2} y cos^{2n-1}(y) (sin(y))’ dy$ and integrating by parts we get: $I_{n+1}= – 1/(2n)+(2n-1)I_n – (2n-1)I_{n+1}$ which is equivalent with: $I_n/n = 2(I_n – I_{n+1})-1/(2n^2)$. Now, the series is equal to $2I_1-pi^2/12=pi^2/6.$