# A problem on finding the limit of the sum

$$u_{n} = \frac{1}{1\cdot n} + \frac{1}{2\cdot(n-1)} + \frac{1}{3\cdot(n-2)} + \dots + \frac{1}{n\cdot1}.$$

Show that, $$\lim_{n\rightarrow\infty} u_n = 0$$.

The only approach I can see is either finding $$nu_{n}$$ or $$(n+1)u_{n}$$ and seeing that:

$$(n+1) \ u_{n+1} = (1+\frac{1}{n}) + (\frac{1}{2} + \frac{1}{n-1}) + \dots + (\frac{1}{n} + 1).$$

• Here's how to use MathJax to format your equations Apr 19 at 13:36
• What is $1.n$ in your series? Maybe is $1 \cdot n$? In that case you can either use the "\cdot" command or simply drop the $1$ Apr 19 at 13:54

\begin{align}u_n&=\sum_{k=1}^n\frac1{k(n+1-k)}\\ &=\frac1{n+1}\sum_{k=1}^n\left(\frac1k+\frac1{n+1-k}\right)\\ &=\frac2{n+1}\sum_{k=1}^n\frac1k\\ &\sim\frac{2\ln n}n\to0. \end{align} Here, we used the asymptotic estimane of the harmonic number: $$H_n\sim\ln n$$, and the fact that $$\lim_{n\to\infty}\frac{\ln n}n=0$$, but there are many other ways to prove that $$\lim \frac1n\sum_{k=1}^n\frac1k=0.$$

• Thank you so much Apr 20 at 0:26

By rewriting it the same way as Anne Bauval did we have that

$$u_n=\frac{2}{n+1}\sum_{j=1}^n\frac{1}{j}.$$

We use the definition of the limit to show that $$u_n\to0$$ as $$n\to\infty$$.

Let $$\varepsilon>0$$. Observe that if $$j\geq\frac{4}{\varepsilon}$$, then $$\frac{1}{j}\leq\frac{\varepsilon}{4}$$. Let $$M$$ by any integer greater than $$\frac{4}{\varepsilon}$$. Note that then

\begin{align*} u_n &=\frac{2}{n+1}\sum_{j=1}^M\frac{1}{j}+\frac{2}{n+1}\sum_{j=M+1}^n\frac{1}{j} \\ &\leq\frac{2}{n+1}\sum_{j=1}^M\frac{1}{j}+\frac{2}{n+1}\sum_{j=M+1}^n\frac{\varepsilon}{4} \\ &\leq\frac{2}{n+1}\sum_{j=1}^M\frac{1}{j}+\frac{\varepsilon}{2} \\ \end{align*}

for all $$n\geq M$$. Choose now $$N>0$$ such that

$$\frac{2}{n+1}\sum_{j=1}^M\frac{1}{j}<\frac{\varepsilon}{2}$$

for all $$n\geq N$$ (note that $$M$$ is fixed so this can be done). Then, for any $$n\geq\max\{N,M\}$$ we have, by combining our estimates, that

$$0\leq u_n\leq\frac{2}{n+1}\sum_{j=1}^M\frac{1}{j}+\frac{\varepsilon}{2}<\varepsilon$$

for all $$n\geq\max\{N,M\}$$. This proves that

$$\lim_{n\to\infty}u_n=0.$$

$$u_n=\sum_{k=1}^n\frac1{k(n+1-k)}$$

Using partial fraction decomposition $$u_n=\frac 1{n+1}\sum_{k=1}^n \frac 1k-\frac 1{n+1}\sum_{k=1}^n \frac 1{k-n-1}$$

Using harmonic numbers $$u_n=\frac{2 }{n+1}H_n$$

Using asymptotics $$u_n=\frac{2 (\log (n)+\gamma)}{n}+O\left(\frac{\log(n)}{n^2}\right)$$

You could re-index the second summation to have a quicker solution.