Does Changing the Starting Point of a Series Affect Its Sum?

[agnimusayoti]

Homework Statement

ML Boas gave an example to find $S_n$; $S$, and $R_n$ of a series that follow this rule:
$$\sum^{\infty}_{2} \frac {2}{n^2-1}$$.
He show that I have to change the formula of the rule before determine $S_n$; $S$, and $R_n$

Relevant Equations

I can catch up the following explanation. But I still don't understand why he choose to change the form rather than use the initial formula.

1. Is it because the initial formula start the series from $n = 2$?
2. If the initial formula is used, can I find $S$, which $$S=\lim_{n\to\infty} \frac{2}{n^2-1}=\frac{2}{\infty}=0$$? Why that answer is different if the formula is changed.

 

[archaic]

The limit you have done does not represent the sum.
$$S=\sum^{\infty}_{n=2} \frac {2}{n^2-1}=\sum^{\infty}_{n=2} u_n$$
$u_n$ is a sequence! $S$ is the sum of all terms of the sequence, from $k=2$ to "$\infty$".
To further illustrate why that can't be a sum; notice that, for any given $n$, we have $u_n>0$. Do you think that you can sum strictly positive numbers and get $0$?

 

[agnimusayoti]

OH! I see. The sequence is getting smaller, but still bigger than 0. So, the limit is wrong. Then, what does the limit represent? Uhm, I have another question, If I want to change the initial series from $n_i=2$ become $n_i=1$ what should I do? Because the book just show that the denumerator is added by 1. I can remember that rule, but don't understand the idea behind the process. Thanks.

 

[archaic]

OH! I see. The sequence is getting smaller, but still bigger than 0. So, the limit is wrong. Then, what does the limit represent?

The limit is not wrong, but what you thought it represents is what's wrong. The limit you have computed is that of the sequence, just not that of the sum of the terms of the sequence.
The sum is given by this expression:
$$\sum_{n=2}^Ku_n=u_2+u_3+u_4+u_5+...+u_{K-2}+u_{K-1}+u_{K}$$
If you make $K\to\infty$, then you'd be summing up infinitely many terms of $u_n$.
The limit you computed is this:
$$\lim_{n\to\infty}u_n=\lim_{n\to\infty}\frac{2}{n^2-1}=0$$
It just tells you that as $n$ grows bigger, the term $u_n$ is getting closer and closer to $0$. It's similar to the limit of a function, think of it like $u_n=f(n)$. When I tell you that $\lim_{x\to\infty}\frac 1x=0$, you don't say the sum of all $f(x)$s is $0$, it's just that as you plug in bigger and bigger $x$s, the function approaches $0$ more and more.

Uhm, I have another question, If I want to change the initial series from $n_i=2$ become $n_i=1$ what should I do? Because the book just show that the denumerator is added by $1$. I can remember that rule, but don't understand the idea behind the process. Thanks.

If you imagine $n'=n-1$, then you see that for $n=2$ we get $n'=1$, and as $n\to\infty$ both are the same. We also have that $n=n'+1$.
$$S=\sum^{\infty}_{n=2}\frac {2}{n^2-1}=\sum^{\infty}_{n'=1}\frac {2}{(n'+1)^2-1}$$

 

[agnimusayoti]

Thanks again. Your explanation is so clear.
1. I become understand that my work (take limit as n approaching $\infty$ to sigma notation) do not find $S$. This is because if I take limit to formula inside the sigma notation, actually I try to find $U_n$ as $n\to\infty$. So, I haven't understand clearly that
$$S=\lim_{n\to\infty} S_n$$ need the formula for summation from $n_i$ to $n_f$. This formula is gotten by expanding the series and try to find the $S_n$ formula.
2. Relation between n' and n. I have to substitute n in it's relation with n', which the newest initial limit for the series. I don't see that I have to "think backward", from n' to n. (Umm, it's difficult to explain what I mean about "think backward". Anyway, now I understand.

Thank you very much!

 

[archaic]

Thank you very much!

Glad I could help!

 

[Mark44]

Homework Statement:: ML Boas gave an example to find $S_n$; $S$, and $R_n$ of a series that follow this rule:
$$\sum^{\infty}_{2} \frac {2}{n^2-1}$$.
He show that I have to change the formula of the rule before determine $S_n$; $S$, and $R_n$
Relevant Equations:: I can catch up the following explanation. But I still don't understand why he choose to change the form rather than use the initial formula.

1. Is it because the initial formula start the series from $n = 2$?
2. If the initial formula is used, can I find $S$, which $$S=\lim_{n\to\infty} \frac{2}{n^2-1}=\frac{2}{\infty}=0$$?

You can't substitute $\infty$ in any arithmetic expression.

Why that answer is different if the formula is changed.

The expressions $S_n$, $S$, and $R_n$ represent, respectively, the sequence of partial sums, the sum of the series, and the remainder $R_n = S - S_n$.
The sequence of partial sums, $S_n$, is the sum of terms up through index n. IOW, $S_n = \sum_{k=2}^n \frac 2 {k^2 - 1}$. The remainder, $R_n$ is the difference between the sum of the series and $S_n$.
I've read through the responses in this thread, but I didn't see anywhere that you had actually found the sum of the series. The key to this problem is to use the technique of partial fractions to decompose $\frac 2 {n^2 - 1}$ into two simpler fractions. After you do that, the resulting series is a telescoping series whose sum can be found fairly easily.

 

[Mark44]

ML Boas gave an example...

He show that..

M. L. Boas is Mary L. Boas, a woman to the best of my knowledge.

 

[agnimusayoti]

You can't substitute $\infty$ in any arithmetic expression.
The expressions $S_n$, $S$, and $R_n$ represent, respectively, the sequence of partial sums, the sum of the series, and the remainder $R_n = S - S_n$.
The sequence of partial sums, $S_n$, is the sum of terms up through index n. IOW, $S_n = \sum_{k=2}^n \frac 2 {k^2 - 1}$. The remainder, $R_n$ is the difference between the sum of the series and $S_n$.
I've read through the responses in this thread, but I didn't see anywhere that you had actually found the sum of the series. The key to this problem is to use the technique of partial fractions to decompose $\frac 2 {n^2 - 1}$ into two simpler fractions. After you do that, the resulting series is a telescoping series whose sum can be found fairly easily.

Yeah, I see. i can decompose the fraction then determine S, Sn and Rn. Thanks!

 

[agnimusayoti]

M. L. Boas is Mary L. Boas, a woman to the best of my knowledge.

Oh my goodness, what a funny mistake. I thought M.L. Boas is a male. Hehe. I agree that she can Made Mathematical for Physics easier than another book.