Solve the problem involving sum of a series

In summary: ##\left(\frac{1}{6} \sum_{r=1}^{n} \frac{1}{r} - \frac{1}{r+3}\right) + \left(\frac{1}{2} \sum_{r=1}^{n} \frac{1}{r+2} - \frac{1}{r+1}\right)##would be the sum of the first two terms and the last two terms would cancel each other out.
  • #1
chwala
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Homework Statement
Show that ##\dfrac{1}{r(r+1)(r+2)} -\dfrac{1}{(r+1)(r+2)(r+3)}= \dfrac{3}{r(r+1)(r+2)(r+3)}##; hence sum the series ## \sum_{r=1}^n \dfrac{1}{r(r+1)(r+2)(r+3)}##
Relevant Equations
Method of difference
1671706716004.png


Attempt;
##\dfrac{1}{r(r+1)(r+2)} -\dfrac{1}{(r+1)(r+2)(r+3)}=\dfrac{(r+3)-1(r)}{r(r+1)(r+2)(r+3)}=\dfrac{3}{r(r+1)(r+2)(r+3)}##

Let ##f(r)=\dfrac{1}{r(r+1)(r+2)}##

##f(r+1)= \dfrac{1}{(r+1)(r+2)(r+3)}##

Therefore ##\dfrac{3}{r(r+1)(r+2)(r+3)}## is of the form ##f(r)-f(r+1)##

When ##r=1## we have

##\left[\dfrac{1}{6}- \dfrac{1}{24}\right]##

When ##r=2## we have

##\left[\dfrac{1}{24}- \dfrac{1}{60}\right]##

When ##r=3## we have

##\left[\dfrac{1}{60}- \dfrac{1}{120}\right]##

...

When ##r=n## we shall have

##\dfrac{1}{n(n+1)(n+2)}- \dfrac{1}{(n+1)(n+2)(n+3)}##

Therefore;

##3\sum_{r=1}^n \dfrac{1}{r(r+1)(r+2)(r+3)}=\left[\dfrac{1}{6}-\dfrac{1}{(n+1)(n+2)(n+3)}\right]##

##\sum_{r=1}^n \dfrac{1}{r(r+1)(r+2)(r+3)}=\left[\dfrac{1}{18}-\dfrac{1}{3(n+1)(n+2)(n+3)}\right]##

Your insight welcome...refreshing on this today...
 
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  • #2
Looks good. But rather than working out the numerical values for a few early terms, a more ‘formal’ proof might be something like this:

Let ##g(r) = \frac 3{r(r+1)(r+2)(r+3)}##. You have already defined ##f(r)## and established that ##g(r) = f(r) – f(r+1)##.

##\begin {flalign*}
\sum_{r=1}^n g(r)
&= \sum_{r=1}^n (f(r)-f(r+1))\\
&= f(1)-f(2)+f(2)-f(3)+f(3)-f(4)...+f(n)-f(n+1)\\
&= f(1)-f(n+1)\\
\end {flalign*}##
etc.

Edit - typo's
 
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  • #3
@chwala ,
In each of your similarly titled threads, you are looking for a result which can be written to depend upon the difference of two sums,
These are: ##\displaystyle \quad \sum_{r=1}^n f(r) \quad ## and ##\displaystyle \quad \sum_{r=1}^n f(r+1) \quad ## .

Notice that the first of these can be written as ##\displaystyle \quad \sum_{r=1}^n f(r) = f(1) + \sum_{r=2}^n f(r) ## .

The second can be written as:

##\displaystyle \sum_{r=1}^n f(r+1) = \sum_{r=2}^{n+1} f(r) ##
##\displaystyle \quad \quad \quad \quad = f(n+1)+ \sum_{r=2}^n f(r) \quad ##
 
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  • #4
I have literally studied the examples in my textbook (solved numerically) and applied the same procedure to the given problem...are you of the opinion that the 'formal' approach is the most concrete way of proof? cheers @SammyS @Steve4Physics
 
  • #5
chwala said:
I have literally studied the examples in my textbook (solved numerically) and applied the same procedure to the given problem...are you of the opinion that the 'formal' approach is the most concrete way of proof? cheers @SammyS @Steve4Physics
I’m not a mathematician but, IMO, it’s a case of rigour and aesthetics.

General proofs should be independent of specific numerical values. It’s worth developing the skill to present proofs in this way.

Incidentally, I like @SammyS's method better than mine.
 
  • #6
Pal @chwala we can directly decompose ##\frac{1}{r (r+1) (r+2) (r+3)}## as
$$
\frac{1/6}{r} - \frac{1/2}{r+1} + \frac{1/2}{r+2} - \frac{1/6}{r+3}
$$
Apply sigma and segregate as follows:
$$
\begin{align*}
\left(\frac{1}{6} \sum_{r=1}^{n} \frac{1}{r} - \frac{1}{r+3}\right) + \left(\frac{1}{2} \sum_{r=1}^{n} \frac{1}{r+2} - \frac{1}{r+1}\right) \\
\end{align*}
$$

The calculations can be carried on...
 
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  • #7
Hall said:
Pal @chwala we can directly decompose ##\frac{1}{r (r+1) (r+2) (r+3)}## as
$$
\frac{1/6}{r} - \frac{1/2}{r+1} + \frac{1/2}{r+2} - \frac{1/6}{r+3}
$$
Apply sigma and segregate as follows:
$$
\begin{align*}
\left(\frac{1}{6} \sum_{r=1}^{n} \frac{1}{r} - \frac{1}{r+3}\right) + \left(\frac{1}{2} \sum_{r=1}^{n} \frac{1}{r+2} - \frac{1}{r+1}\right) \\
\end{align*}
$$

The calculations can be carried on...
Appreciated @Hall , is the decomposition (i.e the part where you are segregating) correct? Let me go through that...
 

1. What is the formula for finding the sum of a series?

The formula for finding the sum of a series is S = (n/2)(a + l), where n is the number of terms in the series, a is the first term, and l is the last term.

2. How do I know if a series is convergent or divergent?

A series is convergent if the limit of its partial sums approaches a finite number as the number of terms increases. It is divergent if the limit of its partial sums approaches infinity or does not exist.

3. Can I use the same formula to find the sum of any series?

No, the formula for finding the sum of a series only applies to arithmetic series, where the difference between consecutive terms is constant. For other types of series, different formulas may be used.

4. How do I determine the number of terms in a series?

The number of terms in a series can be determined by counting the terms or by using a formula. For example, for an arithmetic series, the number of terms can be calculated using the formula n = (l - a)/d + 1, where l is the last term, a is the first term, and d is the common difference.

5. Can I use a calculator or computer program to find the sum of a series?

Yes, there are many online calculators and computer programs available that can help you find the sum of a series. However, it is important to understand the concepts and formulas behind the calculations in order to use these tools effectively.

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