Confused about the difference method

[Enelafoxa]

Homework Statement

Use the diffence method to sum the series $\sum_{n=2}^N\frac {2n-1}{2n^2(n-1)^2}$

Homework Equations

from the textbook by Riley i have $S_N=\sum_{k=1}^m f(N-k+1)-\sum_{k=1}^m f(1-k)$

The Attempt at a Solution

I prefer to write it as $S_N=\sum_{k=1}^m f(N-m+k)-\sum_{k=1}^m f(k-m)$
And i write the problem as $S=-\frac1 2 \sum_{n=0}^{N-2} \left(\frac1{(n+2)^2}-\frac1{(n+1)^2}\right)$ so we have $f(n)=-\frac1{2(n+2)^2}$ and $m=1$

$\begin{align}S_N &=\sum_{k=1}^1 f((N-2)-1+k) - \sum_{k=0}^1 f(k-1)\nonumber\\&=f(N-2)-f(0)\nonumber\\&=-\frac1 2\left(\frac 1 {N^2} - \frac 1 {2^2}\right)\nonumber\end{align}$ I know this is wrong but why?
But if i write the problem as $S=-\frac 1 2 \sum_{n=1}^{N-1} \left(\frac1{(n+1)^2}-\frac 1 {n^2}\right)$ so that $f(n)=-\frac 1{2(n+1)^2}$ and $m=1$. Processing as the previous result we get $S_N=-\frac 1 2\left(\frac 1 {N^2} - 1\right)$ and i get the right answer.
My question is why the first way is wrong and the second way is right ( i mean it produce the right answer). I know how to do this problem without using the formula given by the book. I just want to try it with the difference method.

 

[Ray Vickson]

Homework Statement

Use the diffence method to sum the series $\sum_{n=2}^N\frac {2n-1}{2n^2(n-1)^2}$

Homework Equations

from the textbook by Riley i have $S_N=\sum_{k=1}^m f(N-k+1)-\sum_{k=1}^m f(1-k)$

The Attempt at a Solution

I prefer to write it as $S_N=\sum_{k=1}^m f(N-m+k)-\sum_{k=1}^m f(k-m)$
And i write the problem as $S=-\frac1 2 \sum_{n=0}^{N-2} \left(\frac1{(n+2)^2}-\frac1{(n+1)^2}\right)$ so we have $f(n)=-\frac1{2(n+2)^2}$ and $m=1$

$\begin{align}S_N &=\sum_{k=1}^1 f((N-2)-1+k) - \sum_{k=0}^1 f(k-1)\nonumber\\&=f(N-2)-f(0)\nonumber\\&=-\frac1 2\left(\frac 1 {N^2} - \frac 1 {2^2}\right)\nonumber\end{align}$ I know this is wrong but why?
But if i write the problem as $S=-\frac 1 2 \sum_{n=1}^{N-1} \left(\frac1{(n+1)^2}-\frac 1 {n^2}\right)$ so that $f(n)=-\frac 1{2(n+1)^2}$ and $m=1$. Processing as the previous result we get $S_N=-\frac 1 2\left(\frac 1 {N^2} - 1\right)$ and i get the right answer.
My question is why the first way is wrong and the second way is right ( i mean it produce the right answer). I know how to do this problem without using the formula given by the book. I just want to try it with the difference method.

If your book says $\sum_{k=1}^m f(N-k+1)-\sum_{k=1}^m f(1-k)$ you should throw out the book, because no formula like that can have anything to do with this problem. Are you sure you have not mis-read what the book says?

 

[Enelafoxa]

Yes it is K.F Riley and M.P Hobson' book.. Third edition.. Here they are..

 

[Enelafoxa]

I just curious about the formula... Is it really work? If it is, is there any exception?

 

[Ray Vickson]

Yes it is K.F Riley and M.P Hobson' book.. Third edition.. Here they are..

Well, alright: the formula $\sum_{k=1}^m f(N-k+1)-\sum_{k=1}^m f(1-k)$ can be correct, if you choose $m$ correctly---but you have not said what $m$ should be?

However, it is much better to NOT use formulas unless you understand exactly where they come from. Doing problems from first principles is absolutely the only way to really learn the subject.

Basically, you try to write a sum of the form $\sum_{k=r}^N f(k)$ in the form$\sum_{k=r}^N [g(k) - g(k-r)]$, giving
$$ g(r)-g(0) + g(r+1) - g(1) + \cdots + g(N) - g(N-r).$$
For all large $N$, most of the terms will cancel, leaving only terms in $g(0), g(1), \ldots, g(r-1)$ and $g(N-r+1), \ldots g(N-1), g(N)$. For the special case of $r=1$ this gives $\text{sum} = g(N) - g(0).$

 

[Enelafoxa]

Yes Sir, i already derive it by myself. A little different, but it produce the same thing. This is what i got from my derivation $\sum_{k=1}^m f(N-m+k) - \sum_{k=1}^m f(m-k)$ like i wrote above.. I knew what m mean sir..

 

[Enelafoxa]

Of course i will not using formula that i don't understand the meaning of it, i always trying to reproduce it myself.. As long as i can.. :').

 

[Ray Vickson]

Yes Sir, i already derive it by myself. A little different, but it produce the same thing. This is what i got from my derivation $\sum_{k=1}^m f(N-m+k) - \sum_{k=1}^m f(m-k)$ like i wrote above.. I knew what m mean sir..

OK, so why did you not tell us what $m$ or function $f(n)$ you want to use?

 

[Enelafoxa]

I forgot to wrote it Sir.. Sorry, my bad... :')
$f(n)$ is defined as a function of n such that $u_n=f(n)-f(n-m)$ with m as a 'difference' between the two (i mean like 5-3=2, 5 is differ from 2 by 3, so in this example i have 3 as my 'm'. Sorry i can't be more rigour like a mathemathician because i only have a little knowledge).
For my problem here i have a specific case which the value of m is 1.
Correct me if i wrong sir, i need to study more and more.
Thank you for your attention about my silly question :').

 

[Ray Vickson]

I forgot to wrote it Sir.. Sorry, my bad... :')
$f(n)$ is defined as a function of n such that $u_n=f(n)-f(n-m)$ with m as a 'difference' between the two (i mean like 5-3=2, 5 is differ from 2 by 3, so in this example i have 3 as my 'm'. Sorry i can't be more rigour like a mathemathician because i only have a little knowledge).
For my problem here i have a specific case which the value of m is 1.
Correct me if i wrong sir, i need to study more and more.
Thank you for your attention about my silly question :').

Well,
$$\frac{1}{(n-1)^2}-\frac{1}{n^2} = \frac{2n-1}{n^2 (n^2-1)}$$
for all $n > 1$, so your sum has the form
$$\text{sum} = \frac{1}{2} \sum_{n=2}^N \left[ \frac{1}{(n-1)^2} - \frac{1}{n^2} \right]. $$

 

[Enelafoxa]

That was my first approach to the problem sir.. But there is something bother me.. From that we get $f(n)=-\frac 1 {2n^2}$ and if we use the formula of $S_N$ we get $S_N=f(N)-f(0)$. The problem is the function is undefined at $n=0$. That's why i try to change the original problem.

 

[Ray Vickson]

That was my first approach to the problem sir.. But there is something bother me.. From that we get $f(n)=-\frac 1 {2n^2}$ and if we use the formula of $S_N$ we get $S_N=f(N)-f(0)$. The problem is the function is undefined at $n=0$. That's why i try to change the original problem.

No: look again. The sum goes from $n = 2$ to $n = N$, so $S_N = f(N) - f(1)$. There is no $f(0)$ anywhere.

 

[Enelafoxa]

I understand what you mean sir... But there is something bothering me from the book... For example : Evaluate the sum $$\sum_{n=1}^N \frac 1 {n(n+2)}$$
Using partial fraction we find $$u_n=-\left[\frac 1{2(n+2)}-\frac 1{2n}\right]$$
Hence $u_n=f(n)-f(n-2)$ with $f(n)=-\frac 1 {2(n+2)}$, and so the sum is given by $$\begin{align}S_N&=\sum_{k=1}^m f(N-k+1)-\sum_{k=1}^m f(1-k)\nonumber\\&=\sum_{k=1}^2 f(N-k+1)-\sum_{k=1}^2 f(1-k)\nonumber\\&=f(N)+f(N-1)-f(0)-f(-1)\nonumber\\&=\frac 3 4 -\frac 1 2 \left( \frac 1 {N+2} + \frac 1 {N+1} \right)\nonumber\end{align}$$
There exist $f(-1)$ even if the original summation begin from $n=1$!
So, back to our problem, if i apply the formula from the book directly i'll get $S_N=f(N)-f(0)$ not $S_N=f(N)-f(2)$ as we know it'll produce the right answer. Precisely, my question is why we can't apply the formula directly to produce the right answer? Is there any error in the formula or something i don't urderstand yet? The example above i took from the book. You may look at the picture I've uploaded before.

 

[Ray Vickson]

I understand what you mean sir... But there is something bothering me from the book... For example : Evaluate the sum $$\sum_{n=1}^N \frac 1 {n(n+2)}$$
Using partial fraction we find $$u_n=-\left[\frac 1{2(n+2)}-\frac 1{2n}\right]$$
Hence $u_n=f(n)-f(n-2)$ with $f(n)=-\frac 1 {2(n+2)}$, and so the sum is given by $$\begin{align}S_N&=\sum_{k=1}^m f(N-k+1)-\sum_{k=1}^m f(1-k)\nonumber\\&=\sum_{k=1}^2 f(N-k+1)-\sum_{k=1}^2 f(1-k)\nonumber\\&=f(N)+f(N-1)-f(0)-f(-1)\nonumber\\&=\frac 3 4 -\frac 1 2 \left( \frac 1 {N+2} + \frac 1 {N+1} \nonumber\end{align}$$
There exist $f(-1)$ even if the original summation begin from $n=1$!
So, back to our problem, if i apply the formula from the book directly i'll get $S_N=f(N)-f(0)$ not $S_N=f(N)-f(2)$ as we know it'll produce the right answer. Precisely, my question is why we can't apply the formula directly to produce the right answer? Is ther any error in the formula or something i don't urderstand yet? The example above i took from the book. You may look at the picture I've uploaded before.

We have
$$\frac{1}{n} - \frac{1}{n+2} = \frac{2}{n(n+2)},$$
so
$$S_N = \sum_{n=1}^N \frac{1}{n(n+2)} = \frac{1}{2} \sum_{n=1}^N \left[ \frac{1}{n} - \frac{1}{n+2} \right] $$
This can be written as
$$S_N = \frac{1}{2} \sum_{n=3}^{N+2} \left[ \frac{1}{n-2} - \frac{1}{n} \right]. $$
Think about it carefully.

 

[Enelafoxa]

I think i have my way to get out from this confusion sir..
I see that the author of that book derive the equation from $\sum_{n=1}^N u_n$ he took the summation from $n=1$. So we can use the formula directly only when the summation begin from $n=1$. So i want to generalized this result for the summation that begin with $n=a$.
For $\sum_{n=a}^N u_n$ we define $u_n=f(n)-f(n-m)$
$$\begin{align}S_N&=f(a)-f(a-m)+f(a+1)-f(a+1-m)+f(a+2)-f(a+2-m)+ ... +f(N-1)-f(N-1-m)+f(N)-f(N-m)\nonumber\\&=f(N)+f(N-1)+...+f(N-m+2)+f(N-m+1)-f(a-m)-f(a+1-m)-...-f(a-1)\nonumber\\&=\sum_{k=0}^{m-1} f(N-k) - \sum_{k=0}^{m-1} f(a+k-m)\nonumber\end{align}$$
Now we can apply this directly to our original problem with $a=2$
$$\begin{align}S_N&=\sum_{k=0}^0 f(N-k)-\sum_{k=0}^0 f(1+k)\nonumber\\&=f(N)-f(1)\nonumber\\&=-\frac 1 2 \left(\frac 1{N^2} -1\right)\nonumber\end{align}$$
There i got the right answer finally... :')