C L's sequence

From the question,

$\displaystyle a_n=\sum_{r=1}^n \frac{1}{r}$

$\displaystyle b_n=\sum_{r=1}^n a_r$

$\displaystyle c_n=\sum_{r=1}^n \frac{b_r}{r+1}$

Lets begin with evaluating $b_n$ .

$b_n=a_1+a_2+...+a_n=1+\left(1+\frac{1}{2}\right)+....+\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{n}\right)$

$\displaystyle =n+\frac{n-1}{2}+\frac{n-2}{3}+.....=\sum_{r=1}^n \frac{n-r+1}{r}$

$\displaystyle \Rightarrow b_n=(n+1)\sum_{r=1}^n\frac{1}{r}-\sum_{r=1}^n(1) =(n+1)a_n-n$

Now lets evaluate $c_n$ .

$\displaystyle c_n=\sum_{r=1}^n \frac{b_r}{r+1}=\sum_{r=1}^n \frac{(r+1)a_r-r}{r+1}=\sum_{r=1}^n a_r-\sum_{r=1}^n \frac{r}{r+1}$

The first summation is $b_n$ . The second summation can be evaluated as follows:

$\displaystyle \sum_{r=1}^n \frac{r}{r+1}=\sum_{r=1}^n \left(1-\frac{1}{r+1}\right)=n-a_{n+1}+1$

$\Rightarrow c_n=b_n+a_{n+1}-n-1=(n+1)a_n+a_{n+1}-2n-1$ .

It can be easily seen that

$\displaystyle a_{n+1}-a_n=\frac{1}{n+1} \Rightarrow a_{n}=a_{n+1}-\frac{1}{n+1}$

Therefore, substituting for $a_n$

$\displaystyle c_n=(n+1)a_{n+1}-2n-2+a_{n+1}=(n+2)a_{n+1}-2(n+1)$ .

Put $n=123$ . $\Rightarrow c_{123}=(125)a_{124}-248$ . Hence, $j=125$ and $k=-248$ .

$j-k=125-(-248) =\fbox{373}$

First, we simplify $b_n$ ,

$a_1+a_2+a_3+\dots+a_n$

$=(1)+(1+\frac{1}{2})+(1+\frac{1}{2}+\frac{1}{3})+\dots+(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n})$

$=1(n)+\frac{1}{2}(n-1)+\frac{1}{3}(n-2)+\dots+\frac{1}{n}(1)$

$=\displaystyle \sum_{k=1}^n \Big (\frac{1}{k}(n+1-k)\Big )$

$=\displaystyle \sum_{k=1}^n \Big (\frac{n+1}{k}-1\Big )$

$=\displaystyle \sum_{k=1}^n \Big (\frac{n+1}{k}\Big )-n$

$=(n+1) \displaystyle \sum_{k=1}^n \Big (\frac{1}{k}\Big )-n$

$=(n+1)(a_n)-n$

$=(n+1)(a_{n+1}-\frac{1}{n+1})-n$

$=(n+1)(a_{n+1})-1-n$

$=(n+1)(a_{n+1})-(1+n)$

$=(n+1)(a_{n+1}-1)$

Then, we simplify $c_n$ ,

$\displaystyle \frac{b_1}{2}+\frac{b_2}{3}+\frac{b_3}{4}+\dots+\frac{b_n}{n+1}$

$= \displaystyle \sum_{k=1}^n \Big (\frac{b_k}{k+1}\Big )$

Substitute $b_n=(n+1)(a_{n+1}-1)$ ,

$= \displaystyle \sum_{k=1}^n \Big (\frac{(k+1)(a_{k+1}-1)}{k+1}\Big )$

$= \displaystyle \sum_{k=1}^n \Big (a_{k+1}-1\Big )$

For $c_{123}$ , substitute $n=123$ ,

$\displaystyle \sum_{k=1}^{123} \Big (a_{k+1}-1 \Big )$

$=\displaystyle \sum_{k=1}^{123} \Big (a_{k+1}\Big ) -123$

$=\Big (b_{124}-a_1\Big )-123$

$=b_{124}-124$

Substitute $b_n=(n+1)(a_n)-n$ ,

$=\Big (125a_{124}-124\Big )-124$

$=125a_{124}-248$

Hence, $j=125$ and $k=-248$

$j-k=373$

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Ok I have finished my solution. This part is not very important, just sharing how I get through this problem.

My first attempt is solve for $b_n$ because I think that if I can express $b_n$ as some $a_n$ , it will be very useful when I solve for $c_n$ . For me $b_n$ is like the bridge between $a_n$ and $c_n$ .

When I try to simplify $b_n$ , I stopped at

$b_n=(n+1)(a_n)-n$

Then, I noticed when I simplify $c_n$ and substitute $b_n$ I will get an ugly

$= \displaystyle \sum_{k=1}^n \Big (\frac{(k+1)(a_k)-k}{k+1}\Big )$

So I continue my journey on $b_n$ to find a better expression

$b_n=(n+1)(a_{n+1}-1)$

And finally I can get $c_n$ with the help of both

$b_n=(n+1)(a_{n+1}-1)$

$b_n=(n+1)(a_n)-n$

This means that my first stop is not useless at all!

Then I get a beautiful expression of

$c_{123}=125a_{124}-248$

Observe $b_n = \sum_{k=1}^n a_k = \sum \frac{n+1-k}{k} = -n + (n+1)a_n.$

And $c_n = \sum \frac{b_k}{k+1} = \sum \frac{-k}{k+1} + a_k$

$= -n + b_n + \sum \frac{1}{k+1} = -(2n+1) + (n+1)a_n + a_{n+1} = -(2n+2) + (n+2)a_{n+1}$

The difference is 3n+4. Plugging n=123, answer is $\fbox{373}$ .

Let's try to address the general case instead. Note that $b_{n-1} = \sum_{i=1}^{n-1} a_i$ $= 1+\left (1 + \frac{1}{2} \right ) + \left ( 1 + \frac{1}{2} + \frac{1}{3} \right ) + \cdots + \left ( 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n-1} \right )$ Rearranging the terms, we obtain: $b_{n-1} = n-1 + \frac{n-2}{2} + \frac{n-3}{3} +\cdots + \frac{1}{n-1}$ $= n + \frac{n}{2} + \frac{n}{3} + \cdots + \frac{1}{n-1} - (n-1)$ $= n \left (1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n-1} \right ) - n \times \frac{1}{n} - (n-1)$ $= n \times a_n - n$ Thus, $a_i-1= \frac{a_{i-1}}{i}$ for all $i \in \mathbb{N}$ . Now, note that: $c_n= \sum \limits_{k=2}^{n} \frac{b_{k-1}}{k} = \sum \limits_{k=2}^{n} (a_k - 1)$ $=a_n + b_{n-1} - a_1 - (n-1) = (n+1)a_n - 2n$ Hence, $c_{123} = 125 a_{124} - 248$ . Hence, we have $i= 125$ , $j= 248$ , $i-j= \boxed{373}$ .

Lemma 1 . $\frac{b_{n}}{n + 1} - \frac{b_{n - 1}}{n} = \frac{1}{n + 1}$

Proof . $\frac{b_{n}}{n + 1} - \frac{b_{n - 1}}{n} = \frac{nb_{n} - (n + 1)b_{n - 1}}{n(n + 1)}$ . We now compute $nb_{n} - (n + 1)b_{n - 1}$ .

$nb_{n} - (n + 1)b_{n - 1} = n(a_{1} + a_{2} + ... + a_{n}) - (n + 1)(a_1 + a_2 + ... + a_{n - 1}) = na_{n} - a_1 - a_2 - a_{n - 1} = (a_n) + (a_n - a_1) + ... + (a_n - a_{n - 1}) = 1 \cdot \frac{1}{1} + 2 \cdot \frac{1}{2} + ... + n \cdot \frac{1}{n} = n$ .

This proves the lemma.

Lemma 2. $b_n = (n + 1)a_n - n$

Proof. $b_n = \sum^{n}_{i = 1}(\frac{1}{i}(n + 1 - i)) = \sum^{n}_{i = 1}(\frac{n + 1}{i} - 1) = (n + 1)(1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}) - n = (n + 1)a_n - n$ . This proves the lemma.

Lemma 3 . $c_n = b_{n + 1} - n - 1$

Proof : $\frac{b_n}{n + 1} = \frac{b_{n - 1}}{n} + \frac{1}{n + 1} = \frac{b_{n - 2}}{n - 1} + \frac{1}{n} + \frac{1}{n + 1} = ... = \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n + 1} = a_{n + 1} - 1$ . (By repeated application of Lemma 1.)

$c_n = \frac{b_1}{2} + \frac{b_2}{3} + ... + \frac{b_n}{n + 1} = a_2 - 1 + a_3 - 1 + ... + a_{n + 1} - 1 = b_{n + 1} - n - 1$ . This proves the lemma.

Lemma 4 . $c_n = (n + 2)a_{n + 1} - (2n + 2)$ .

Proof : By Lemma $2$ and $3$ ,

$c_n = b_{n + 1} - n - 1 = (n + 2)a_{n + 1} - n - 1 - n - 1 = (n + 2)a_{n + 1} - (2n + 2)$ . This proves the final lemma.

By Lemma $4$ , $j = 125, k = -248$ satisfies the condition.

So, $j - k = \boxed{125 - (-248) = 373}$ .

Note : However, I still can't figure out why this is the only answer. I just assume that this is unique from the problem statement.

Start with $b_n$ , $b_n=\sum_{i=1}^{n}a_i=1+(1+\frac{1}{2})+(1+\frac{1}{2}+\frac{1}{3})+\dots+(1+\frac{1}{2}+\dots +\frac{1}{n}),$ we notice that there are $n$ 1's, $(n-1)$ 2's, $(n-2)$ 3's, etc. So $b_n=\sum_{i=1}^{n}{1}+\sum_{i=1}^{n-1}{\frac{1}{2}}+\sum_{i=1}^{n-2}{\frac{1}{3}}+\dots+\sum_{i=1}^{2}{\frac{1}{n-1}}+\sum_{i=1}^{1}{\frac{1}{n}}$

$b_n=n+\frac{n-1}{2}+\frac{n-2}{3}+\dots+\frac{2}{n-1}+\frac{1}{n}$ $\boxed{b_n=\sum_{i=1}^{n}{\frac{(n+1)-i}{i}}}$

we found a very easy expression for $b_n$ .\

Now we are going to work with $c_n$ $c_n=\sum_{m=1}^{n}{\frac{b_m}{m+1}},$ $c_n=\sum_{m=1}^{n}{\frac{\sum_{i=1}^{m}{\frac{(m+1)-i}{i}}}{m+1}}$ $c_n=\sum_{m=1}^{n}{\sum_{i=1}^{m}{\frac{(m+1)-i}{i(m+1)}}},$ is easy to see that $\frac{(m+1)-i}{i(m+1)}=\frac{1}{i}-\frac{1}{m+1}$ $c_n=\sum_{m=1}^{n}{\sum_{i=1}^{m}{(\frac{1}{i}-\frac{1}{m+1})}},$ $c_n=\sum_{m=1}^{n}{(\sum_{i=1}^{m}{\frac{1}{i}}-\sum_{i=1}^{m}{\frac{1}{m+1})}},$ $c_n=\sum_{m=1}^{n}{(a_m-\frac{m}{m+1})},$ we see that $-\frac{m}{m+1}=\frac{1}{m+1}-1,$ hence $c_n=\sum_{m=1}^{n}{(a_m+\frac{1}{m+1}-1)},$ but $a_m+\frac{1}{m+1}=a_{m+1},$ $c_n=\sum_{m=1}^{n}{(a_{m+1}-1)},$ $c_n=\sum_{m=1}^{n}{a_{m+1}}-\sum_{m=1}^{n}{1},$

$\boxed{c_n=\sum_{m=1}^{n}{a_{m+1}}-n}$ At this point we are going to write the $\sum_{m=1}^{n}{a_{m+1}}$ depending on $b_n$ ' $s$ to get a closed form. $\sum_{m=1}^{n}{a_{m+1}}=a_2+a_3+\dots+a_n+a_{n+1},$ add and substract $a_1$ , $\sum_{m=1}^{n}{a_{m+1}}={\color{#D61F06} {a_1} }+a_2+a_3+\dots+a_n+a_{n+1}-{\color{#D61F06} {a_1} },$ we have that $a_1=b_1$ hence $\sum_{m=1}^{n}{a_{m+1}}=b_{n+1}-b_1,$ now go back to $c_n$ $\boxed{c_n=b_{n+1}-b_1-n}$

Substituting $b_{n+1}$ y $b_1$ $c_n=\sum_{i=1}^{n+1}{\frac{((n+1)+1)-i}{i}}-1-n$ $c_n=\sum_{i=1}^{n+1}{(\frac{n}{i}+\frac{2}{i}-\frac{i}{i})}-(n+1)$ $c_n=\sum_{i=1}^{n+1}{(n\frac{1}{i}+2\frac{1}{i}-1)}-(n+1)$ $c_n=n\sum_{i=1}^{n+1}{\frac{1}{i}}+2\sum_{i=1}^{n+1}{\frac{1}{i}}-\sum_{i=1}^{n+1}{1}-(n+1)$ $c_n=na_{n+1}+2a_{n+1}-(n+1)-(n+1)$ $\boxed{c_n=(n+2)a_{n+1}-2(n+1)}.$ we're done. Let $j=n+2, k=2(n+1)$ and $n=123$ , so $c_{123}=(123+2)a_{123+1}-2(123+1)$ $\boxed{c_{123}=125a_{124}-248}$ in this case $j=125, k=-248,$ so $\boxed{j-k=373}$ .