$u_n=\frac {1}{1\cdot n}+\frac {1}{2(n-1)}+\ldots+\frac {1} {n \cdot 1}$,then $\lim u_n=?$

Question

I am stuck on the following problem and do not know to tackle it:

If $u_n=\dfrac {1}{1\cdot n}+\dfrac {1}{2 \cdot (n-1)}+\dfrac {1}{3 \cdot (n-2)} + \ldots +\dfrac {1}{n\cdot1}$, then $\lim u_n = 0$.

Can someone point me in the right direction?

Answer 1

Hint: $(n+1)u_n=(1+\frac {1}{n})+(\frac {1}{2} +\frac {1}{n-1})+....+(\frac {1}{n}+1)$

Answer 2

This is not the most direct way (definitely not as elegant as learner's solution) but just another possible route.

First show that $u_n$ is bounded above by $1$ and below by $0$. Both are fairly straightforward. To show the upper bound, note that each term in $u_n$ is $\leq \dfrac1n$ (Why?).

Now proceed as follows \begin{align} u_n & = \sum_{k=1}^n \dfrac1{k(n+1-k)} = \sum_{k=1}^n \left(\int_0^1 x^{k-1} dx \right) \left(\int_0^1 y^{n-k} dy \right) = \int_{[0,1]^2} y^{n-1} \sum_{k=1}^n \left(\dfrac{x}y \right)^{k-1} dx dy\\ & = \int_{[0,1]^2} \dfrac{y^n-x^n}{y-x} dx dy \end{align} Now take the limit, swap the integral and summation (Why?) and conclude what you want.