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

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:

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

Please help me understand how I can solve this problem using this, or any method of your preference! Thank you so much!!

Aig
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Abhisek Mukherjee
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4 Answers4

6

$$\begin{align}u_n&=\sum_{k=1}^n\frac1{k(n+1-k)}\\ &=\frac1{n+1}\sum_{k=1}^n\left(\frac1k+\frac1{n+1-k}\right)\\ &=\frac2{n+1}\sum_{k=1}^n\frac1k\\ &\sim\frac{2\ln n}n\to0. \end{align}$$ 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 $$\lim \frac1n\sum_{k=1}^n\frac1k=0.$$

Anne Bauval
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4

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

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

Lorago
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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.

Claude Leibovici
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0

The problem can be generalized. Assume $ a_n\searrow 0$ and $\sum a_n^2<\infty.$ Let $$u_n=\sum_{k=1}^na_ka_{n-k}$$ Fix $N$ and let $n\ge 2N.$Then $$u_n\le \sum_{k=1}^Na_ka_{n-k}+\sum_{k=n-N+1}^na_ka_{n-k}+\sum_{k=N+1}^{n-N} a_ka_{n-k}\\ \le 2Na_1a_{n-N}+{1\over 2}\sum_{k=N+1}^{n-N}[a_k^2+a_{n-k}^2] \le 2Na_1a_{n-N}+\sum_{k=N}^\infty a_k^2$$ The estimate implies $\limsup u_n=\sum_N^\infty a_k^2.$ As $N$ is arbitrary we get $\lim u_n=0.$ We can apply the above to $a_n=n^{-\alpha} $ for $\alpha>{1\over 2}.$

Ryszard Szwarc
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