# Solve the problem involving sum of a series

• chwala
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.

#### chwala

Gold Member
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

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...

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

chwala
@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,

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) ##

Steve4Physics and chwala
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

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.

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...

Greg Bernhardt
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...