# if $\lim\limits_{n \to \infty} b_n =0$ then how to prove that $\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}=0$

in Problems in Mathematical Analysis I problem 2.3.16 a),

if $$\lim\limits_{n \to \infty}a_n =a$$, then find $$\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}$$

The proof that was on the answers:- by Toeplitz theorem let $$c_{n,k} =\frac{1}{(n+1-k)(n+2-k)}$$ and the answer is $$a$$, here I didn't understand how this is possible because one condition for Toeplitz theorem is $$\lim\limits_{n \to \infty} c_{n,k} =0 \ \forall k \in \mathbb{N}$$ doesn't hold if $$k=n$$ for example

$$\textbf{Toeplitz theorem :}$$ let $$\{ c_{n,k}: 1\leq k \leq n \}$$ be an array of real numbers such that:

1-$$\lim\limits_{n \to \infty} c_{n,k} =0 \ \forall k \in \mathbb{N}$$

2- $$\sum\limits_{k=1} ^{n} c_{n,k} \to 1$$ as $$n\to \infty$$

3- There is exist $$C \in \mathbb{R}$$ such that for all positive integer $$n$$ $$\sum\limits_{k=1} ^{n} |c_{n,k}| \leq C$$

then for any converging sequence $$\{a_n \}$$ the sequence $$\{ b_n \}$$ given by $$b_n:=\sum\limits_{k=1} ^{n} c_{n,k} a_k$$ is convergent and $$\lim\limits_{n \to \infty}b_n= \lim\limits_{n \to \infty}a_n$$

$$\textbf{ My Proof:-}$$

First I have to prove that the sequence $$\{ b_n\}$$ is convergent this can be shown by considering the sum $$s_n =\sum\limits_{k=1} ^{n} |c_{n,k} a_k|$$ since the terms of the sum consist only positive number and since $$a_n$$ is convergent sequence then there exist some $$M \in \mathbb{R}$$ such that $$a_k\leq M \ \forall k$$ then the sequence $$s_n then the sequence $$s_n$$ converge then $$b_n$$ converge Now lets proof that $$\lim\limits_{n \to \infty}b_n= \lim\limits_{n \to \infty}a_n$$ first let $$a:=\lim\limits_{n \to \infty}a_n$$ since $$\sum\limits_{k=1} ^{n} c_{n,k} \to 1$$ as $$n\to \infty$$ then I need to prove that $$\sum\limits_{k=1} ^{n} c_{n,k} a_k-a \sum\limits_{k=1} ^{n} c_{n,k} \to 0 \text{ as } n \to \infty$$ $$S_n:=\sum\limits_{k=1} ^{n} c_{n,k} (a_k-a)$$ $$\text{since } a_k \to a \text{ then } \forall \varepsilon_1 >0 \ \exists N \in \mathbb{N} \text{ such that } \forall n \geq N \ |a_k-a|<\varepsilon_1$$ $$\text{Choose \varepsilon =\inf \{ \varepsilon_1 , \frac{\varepsilon_1}{ C M}\} } \text{Choose n such that } \sup\{ |c_{n,k}| \} < \frac{\varepsilon_1}{NM}$$ $$-\varepsilon - \sum\limits_{k=N+1} ^{n} |c_{n,k} a_k| \leq S_n \leq \varepsilon + \sum\limits_{k=N+1} ^{n} |c_{n,k} a_k|$$ since $$\sum\limits_{k=N+1} ^{n} |c_{n,k} a_k| <\varepsilon$$ then $$-2\varepsilon

I tried to prove it myself and this what I got

let $$b_{n-1}= a_n - a_{n-1}$$ it is easy to see that $$\lim\limits_{n \to \infty }b_n =0$$ (because any converging sequence in $$\mathbb{R}$$ is a Cauchy sequence) so $$\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}=\lim\limits_{n \to \infty} \left( a_n - \frac{a_1}{n+1} +b_n -\sum\limits_{k =1} ^n \frac{b_k}{n+1-k} \right)=a -\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}$$ and here I couldn't prove that $$\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k} =0$$

I also want to ask why is the answers that the book provide is correct ? because one of the necessary condition of Toeplitz theorem doesn't hold

EDIT

@TheSilverDoe has an excellent answer but now I don't know where my mistake was since $$\lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k} \neq 0$$ and $$\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}= \lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)}- \frac{a_k}{(n+2-k)}$$ $$=\lim\limits_{n \to \infty}+a_n -\frac{a_1}{n+1}- \sum\limits_{k=1 }^{n-1} \frac{a_{n+1-k}- a_{n-k}}{k+1}$$ $$= a-\lim\limits_{n \to \infty} \sum_{k=1}^{n-1}\frac{a_{k+1}- a_{k}}{n+1-k}=a-\lim\limits_{n \to \infty} \sum_{k=1}^{n-1}\frac{b_{k}}{n+1-k}=a-\lim\limits_{n \to \infty} \sum_{k=1}^{n}\frac{b_{k}}{n+1-k}- b_{n}=$$ $$a- \lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}$$

I still can't figure out where is the mistake in my logic.

Is the mistake related to conditional convergent series and Riemann rearrangement theorem ?

## 2 Answers

First part : The statement in the title is wrong.

Let's suppose that the sequence $$(b_k)_{k \geq 1}$$ is the following : $$\dfrac{1}{\ln(2)}, \dfrac{1}{\ln(2)}, \dfrac{1}{\ln(3)}, \dfrac{1}{\ln(3)}, \dfrac{1}{\ln(3)}, \dfrac{1}{\ln(4)}, \dfrac{1}{\ln(4)}, \dfrac{1}{\ln(4)}, \dfrac{1}{\ln(4)}, ...$$

(the $$\dfrac{1}{\ln(k)}$$ term appearing $$k$$ times).

Let $$N_1=0$$ and for any $$j \geq 2$$, let $$N_j = 2+3+...+j$$. Then \begin{align*}\sum_{k=1}^{N_j} \dfrac{b_k}{N_j+1-k} &= \sum_{i=2}^j \sum_{k=N_{i-1}+1}^{N_i} \dfrac{b_k}{N_j+1-k} \\ &= \sum_{i=2}^j \sum_{k=N_{i-1}+1}^{N_i} \dfrac{1}{\ln(i)(N_j+1-k)} \\ &= \sum_{i=2}^j \dfrac{1}{\ln(i)}\sum_{k=N_{i-1}+1}^{N_i} \dfrac{1}{N_j+1-k} \\ &\geq \dfrac{1}{\ln(j)}\sum_{i=2}^j \sum_{k=N_{i-1}+1}^{N_i} \dfrac{1}{N_j+1-k} \\ &=\dfrac{1}{\ln(j)}\sum_{k=1}^{N_j} \dfrac{1}{N_j+1-k} \\ &=\dfrac{1}{\ln(j)}\sum_{k=1}^{N_j} \dfrac{1}{k} \\ &\geq \dfrac{1}{\ln(j)}\sum_{k=1}^{j} \dfrac{1}{k} \end{align*}

Since $$\displaystyle\lim_{j \rightarrow +\infty} \dfrac{1}{\ln(j)}\sum_{k=1}^{j} \dfrac{1}{k} = 1$$, then $$\displaystyle\sum_{k=1}^{N_j} \dfrac{b_k}{N_j+1-k}$$ does not tend to $$0$$ as $$j$$ tends to $$+\infty$$.

Therefore, $$\displaystyle\sum_{k=1}^{n} \dfrac{b_k}{n+1-k}$$ does not tend to $$0$$ as $$n$$ tends to $$+\infty$$.

Second part : Lets' prove that if $$\displaystyle\lim_{n \rightarrow +\infty} a_n=a$$, then $$\displaystyle\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)} = a$$.

Let's suppose that $$\displaystyle\lim_{n \rightarrow +\infty} a_n=a$$, and let $$\varepsilon > 0$$.

By definition of convergence, there exists $$n_0 \in \mathbb{N}$$ such that for every $$n > n_0$$, one has $$|a_n-a| \leq \varepsilon$$.

Let $$M=\max\lbrace |a_0-a|, ..., |a_{n_0}-a| \rbrace$$.

Finally, let $$N \geq n_0$$ such that for every $$n \geq N$$, $$\dfrac{|a|}{n+1} + M \left[\frac{1}{n+1-n_0}-\frac{1}{n+1}\right] \leq \varepsilon$$

For every $$n > 0$$, one has $$\sum\limits_{k=1 }^n \frac{a}{(n+1-k)(n+2-k)} = a \sum\limits_{k=1 }^n \frac{1}{n+1-k}-\frac{1}{n+2-k} = a \left(1-\dfrac{1}{n+1}\right)$$

so $$a=\dfrac{a}{n+1} +\sum\limits_{k=1 }^n \frac{a}{(n+1-k)(n+2-k)}$$

One deduces that for every $$n \geq N$$, \begin{align*} &\left|a-\sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}\right| \\ \leq &\left|\dfrac{a}{n+1} + \sum\limits_{k=1 }^n \frac{a-a_k}{(n+1-k)(n+2-k)}\right| \\ \leq &\dfrac{|a|}{n+1} + \sum\limits_{k=1 }^n \frac{|a_k-a|}{(n+1-k)(n+2-k)} \\ \leq &\dfrac{|a|}{n+1} + \sum\limits_{k=1 }^{n_0} \frac{|a_k-a|}{(n+1-k)(n+2-k)}+ \sum\limits_{k=n_0+1 }^{n} \frac{|a_k-a|}{(n+1-k)(n+2-k)} \\ \leq &\dfrac{|a|}{n+1} + \sum\limits_{k=1 }^{n_0} \frac{M}{(n+1-k)(n+2-k)}+ \sum\limits_{k=n_0+1 }^{n} \frac{\varepsilon}{(n+1-k)(n+2-k)} \\ \leq &\dfrac{|a|}{n+1} + M \sum\limits_{k=1 }^{n_0} \left[\frac{1}{n+1-k}-\frac{1}{n+2-k}\right]+ \varepsilon\sum\limits_{k=n_0+1 }^{n} \left[\frac{1}{n+1-k}-\frac{1}{n+2-k}\right] \\ \leq &\dfrac{|a|}{n+1} + M \left[\frac{1}{n+1-n_0}-\frac{1}{n+1}\right]+ \varepsilon \left[1-\frac{1}{n+1-n_0}\right] \\ \leq & \varepsilon + \varepsilon \leq 2 \varepsilon \end{align*}

which finally proves that $$\boxed{\displaystyle\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)} = a}$$.

• What about the statement $\lim\limits_{n \to \infty} \frac 1n \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}=0$ ? Nov 22, 2023 at 13:25
• It would be true if I am not mistaken. I guess that one can prove $$\lim\limits_{n \to \infty} \frac{1}{\ln(n)} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}=0$$ Nov 22, 2023 at 13:39
• so that $\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)} =a$ is not correct , right ? and the answers was wrong about Toeplitz theorem
– pie
Nov 22, 2023 at 14:54
• @pie No, the statement with $a_k$ seems to be true ; I added a proof. Nov 22, 2023 at 15:37
• WOW!!! that was an amazing proof Thank you very much Sir
– pie
Nov 22, 2023 at 15:45

I made a mistake that is $$b_n \to 0$$ is true but there is another condition $$s_n :=\sum _{k=1}^n b_k = \sum _{k=1}^n a_{k+1}- a_k= a_{n+1 }- a_1$$ it is clear that $$s_n \to a -a_1$$ i.e $$\displaystyle\sum _{k=1}^\infty b_k$$ converge, in the example of @TheSilverDoe $$\displaystyle\sum _{k=1}^\infty b_k$$ diverge.

$$\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}= a- \lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}$$

define $$\displaystyle L_n:= \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}$$

since $$\displaystyle\sum _{k=1}^\infty b_k$$ converge then $$\forall \varepsilon>0$$ there is exist some $$N \in \mathbb{N}$$ st that $$\forall m_1,m_2 \ge N , m_2>n_1$$, $$\displaystyle -\frac{\varepsilon}{2}<\sum _{k=m_1}^{m_2} b_k <\frac{\varepsilon}{2}$$ (Cauchy sequence )

let $$\displaystyle M =\sup_{k=1} ^{\infty } \{|b_k| \}$$

choose $$\displaystyle n_1$$ st $$\displaystyle\frac{MN}{n_1-N} < \frac{\varepsilon}{2}$$ and let $$n = \sup\{ n_1 , 2N \}$$

$$L_n = \sum\limits_{k =1} ^{N-1} \frac{b_k}{n+1-k} +\sum\limits_{k =N} ^{n} \frac{b_k}{n+1-k }$$

since $$\left|\displaystyle \sum\limits_{k =N} ^{n} {b_k}\right|< \frac{\varepsilon}{2}$$, $$\displaystyle \left|\sum\limits_{k =N} ^{n} \frac{b_k}{n+1-k }\right| <\frac{\varepsilon}{2}$$

$$-\varepsilon

so $$\displaystyle L_n \to 0$$ that proves:- $$\lim\limits_{n \to \infty} \sum\limits_{k=1 }^n \frac{a_k}{(n+1-k)(n+2-k)}= a- \lim\limits_{n \to \infty} \sum\limits_{k =1} ^n \frac{b_k}{n+1-k}= a$$