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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 <M , \sum\limits_{k=1} ^{n} |c_{n,k}| \leq C M $ 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 $$-2\varepsilon<S_n< 2\varepsilon$$


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 ?

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2 Answers 2

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

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

$$-\varepsilon <L_n <\varepsilon $$

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

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