Question

Find the sum of the following series $\dfrac{1}{\left| \!{\underline {\,
  r \,}} \right. }+\dfrac{\left| \!{\underline {\,
  2 \,}} \right. }{\left| \!{\underline {\,
  r+1 \,}} \right. }+\dfrac{\left| \!{\underline {\,
  3 \,}} \right. }{\left| \!{\underline {\,
  r+2 \,}} \right. }+....$

Answer 1

Hint: We write the above series in a general form. Then we will simplify the general expression that we got. We try to get the expression in the form of factorial, this can be done by adding terms or multiplying and dividing with any term. Once we are done simplifying, we will now calculate 1st term, 2nd term ,3rd term and look for the pattern. Then we will get the final term as our answer.

Complete step-by-step answer:
We have, $\dfrac{1}{\left| \!{\underline {\,
  r \,}} \right. }+\dfrac{\left| \!{\underline {\,
  2 \,}} \right. }{\left| \!{\underline {\,
  r+1 \,}} \right. }+\dfrac{\left| \!{\underline {\,
  3 \,}} \right. }{\left| \!{\underline {\,
  r+2 \,}} \right. }+....$
We can write the above series as,
$\Rightarrow {{S}_{n}}=\sum\limits_{n=1}^{n}{\dfrac{n!}{\left( r+n-1 \right)!}}$
Where n=1, 2, 3, 4…….
We will now simplify the above expression.
We know that $n!=n(n-1)!$
Similarly,
$\begin{align}
  & \Rightarrow \sum\limits_{n=1}^{n}{\dfrac{n!}{\left( r+n-1 \right)!}} \\
 & \Rightarrow {{S}_{n}}=\sum\limits_{n=1}^{n}{\dfrac{n!}{\left( r+n-2 \right)!}\times \dfrac{1}{\left( r+n-1 \right)}} \\
\end{align}$
We will now multiply and divide the RHS by (r-2), we will get,
\[\Rightarrow {{S}_{n}}=\sum\limits_{n=1}^{n}{\dfrac{n!}{\left( r+n-2 \right)!}\times \dfrac{r-2}{\left( r+n-1 \right)}}\times \dfrac{1}{r-2}\]
Will now add and subtract n in numerator of expression \[\dfrac{r-2}{\left( r+n-1 \right)}\], we will get,
\[\Rightarrow {{S}_{n}}=\sum\limits_{n=1}^{n}{\dfrac{n!}{\left( r+n-2 \right)!}\times \dfrac{n+r-2-n}{\left( r+n-1 \right)}}\times \dfrac{1}{r-2}\]
We can rewrite the expression \[\dfrac{n+r-2-n}{\left( r+n-1 \right)}\] as,
\[\Rightarrow \dfrac{(n+r-1)-(n+1)}{\left( r+n-1 \right)}\]
We will now simplify the above expression,
\[\Rightarrow 1-\dfrac{(n+1)}{\left( r+n-1 \right)}\]
\[\begin{align}
  & \Rightarrow (r-2){{S}_{n}}=\sum\limits_{n=1}^{n}{\dfrac{n!}{\left( r+n-2 \right)!}\times \left( 1-\dfrac{(n+1)}{\left( r+n-1 \right)} \right)} \\
 & \Rightarrow \sum\limits_{n=1}^{n}{\dfrac{n!}{\left( r+n-2 \right)!}-}\dfrac{n!(n+1)}{\left( r+n-1 \right)\left( r+n-2 \right)!} \\
 & \Rightarrow \sum\limits_{n=1}^{n}{\dfrac{n!}{\left( r+n-2 \right)!}-}\dfrac{(n+1)!}{\left( r+n-1 \right)!} \\
\end{align}\]
We know that, \[{{S}_{n}}={{S}_{1}}+{{S}_{2}}+{{S}_{3}}+{{S}_{4}}.........+{{S}_{n}}\]
We will n=1 in expression \[\dfrac{n!}{\left( r+n-2 \right)!}-\dfrac{(n+1)!}{\left( r+n-1 \right)!}\], we will get
$\begin{align}
  & \Rightarrow (r-2){{S}_{1}}=\dfrac{1}{(r-1)!}-\dfrac{2}{r!} \\
 & \Rightarrow {{S}_{1}}=\dfrac{1}{(r-2)}\times \left( \dfrac{1}{(r-1)!}-\dfrac{2}{r!} \right) \\
\end{align}$
Similarly we can get S1,S2,S3..
$\begin{align}
  & \Rightarrow {{S}_{2}}=\dfrac{1}{(r-2)}\times \left( \dfrac{2!}{r!}-\dfrac{3!}{\left( r+1 \right)!} \right) \\
 & \Rightarrow {{S}_{3}}=\dfrac{1}{(r-2)}\times \left( \dfrac{3!}{\left( r+1 \right)!}-\dfrac{3!}{\left( r+2 \right)!} \right) \\
 & \Rightarrow {{S}_{4}}=\dfrac{1}{(r-2)}\times \left( \dfrac{4!}{(r+2)!}-\dfrac{5!}{\left( r+3 \right)!} \right) \\
\end{align}$
We will put the values in this equation \[{{S}_{n}}={{S}_{1}}+{{S}_{2}}+{{S}_{3}}+{{S}_{4}}.........+{{S}_{n}}\], we will get,
$\begin{align}
  & \Rightarrow {{S}_{n}}=\dfrac{1}{(r-2)}\times \left( \dfrac{1}{(r-1)!}-\dfrac{(n+1)!}{\left( n+r-1 \right)!} \right) \\
 & \Rightarrow {{S}_{n}}=\left( \dfrac{1}{(r-2)!}-\dfrac{(n+1)!}{(r-2)\left( n+r-1 \right)!} \right) \\
\end{align}$

Note: The student makes mistakes while expressing the series in general term. It is important to expand the series by putting the value of n=1,2,3,4….. in the general expression. Students tend to overlook minute things while forming general expressions, so take care while forming the general term.