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Find the sum of the first $n$ terms and up to infinite terms of the following series:
$\dfrac{4}{1\cdot 2\cdot 3}+\dfrac{5}{2\cdot 3\cdot 4}+\dfrac{6}{3\cdot 4\cdot 5}+..................$

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Hint: For solving this question first we will analyse the given series and write the expression of ${{r}^{th}}$ the term of the given series and then write it in the form of ${{T}_{r}}=f\left( r \right)-f\left( r+1 \right)$. After that we will apply the summation and find the sum of first $n$ terms and then we will find the sum of the infinite terms of the given series.

Complete step-by-step answer:
Given:
We have to find the sum of the first $n$ terms and up to infinite terms of the following series:
$\dfrac{4}{1\cdot 2\cdot 3}+\dfrac{5}{2\cdot 3\cdot 4}+\dfrac{6}{3\cdot 4\cdot 5}+..................$
Now, if we look at the given series and if ${{T}_{r}}$ is the ${{r}^{th}}$ term of the given series and try to analyse the first three terms then we can write ${{T}_{r}}=\dfrac{r+3}{r\left( r+1 \right)\left( r+2 \right)}$ .
Now, we have to find the value of $\sum\limits_{r=1}^{n}{{{T}_{r}}}$ . But before this, we should modify the expression of the ${{r}^{th}}$ term with simple arithmetic operations. Then,
$\begin{align}
  & {{T}_{r}}=\dfrac{r+3}{r\left( r+1 \right)\left( r+2 \right)}=\dfrac{r}{r\left( r+1 \right)\left( r+2 \right)}+\dfrac{3}{r\left( r+1 \right)\left( r+2 \right)} \\
 & \Rightarrow {{T}_{r}}=\dfrac{1}{\left( r+1 \right)\left( r+2 \right)}+\dfrac{3}{2}\left[ \dfrac{2}{r\left( r+1 \right)\left( r+2 \right)} \right]=\dfrac{\left( r+2 \right)-\left( r+1 \right)}{\left( r+1 \right)\left( r+2 \right)}+\dfrac{3}{2}\left[ \dfrac{\left( r+2 \right)-r}{r\left( r+1 \right)\left( r+2 \right)} \right] \\
 & \Rightarrow {{T}_{r}}=\dfrac{1}{\left( r+1 \right)}-\dfrac{1}{\left( r+2 \right)}+\dfrac{3}{2}\left[ \dfrac{1}{r\left( r+1 \right)}-\dfrac{1}{\left( r+1 \right)\left( r+2 \right)} \right] \\
 & \Rightarrow {{T}_{r}}=\dfrac{1}{\left( r+1 \right)}+\dfrac{3}{2r\left( r+1 \right)}-\left[ \dfrac{1}{\left( r+2 \right)}+\dfrac{3}{2\left( r+1 \right)\left( r+2 \right)} \right] \\
 & \Rightarrow {{T}_{r}}=\dfrac{2r+3}{2r\left( r+1 \right)}-\dfrac{2r+5}{2\left( r+1 \right)\left( r+2 \right)} \\
\end{align}$
Now, let $f\left( r \right)=\dfrac{2r+3}{2r\left( r+1 \right)}$ . Then,
$\begin{align}
  & f\left( r \right)=\dfrac{2r+3}{2r\left( r+1 \right)} \\
 & \Rightarrow f\left( r+1 \right)=\dfrac{2\left( r+1 \right)+3}{2\left( r+1 \right)\left( r+1+1 \right)} \\
 & \Rightarrow f\left( r+1 \right)=\dfrac{2r+5}{2\left( r+1 \right)\left( r+2 \right)} \\
\end{align}$
Now, as we have calculated that ${{T}_{r}}=\dfrac{2r+3}{2r\left( r+1 \right)}-\dfrac{2r+5}{2\left( r+1 \right)\left( r+2 \right)}$ . Then,
$\begin{align}
  & {{T}_{r}}=\dfrac{2r+3}{2r\left( r+1 \right)}-\dfrac{2r+5}{2\left( r+1 \right)\left( r+2 \right)} \\
 & \Rightarrow {{T}_{r}}=f\left( r \right)-f\left( r+1 \right) \\
\end{align}$
Now, we will calculate the value of $\sum\limits_{r=1}^{n}{{{T}_{r}}}$ . Then,
\[\begin{align}
  & \sum\limits_{r=1}^{n}{{{T}_{r}}}=\sum\limits_{r=1}^{r=n}{f\left( r \right)-f\left( r+1 \right)} \\
 & \Rightarrow \sum\limits_{r=1}^{n}{{{T}_{r}}}=\left[ f\left( 1 \right)-f\left( 2 \right) \right]+\left[ f\left( 2 \right)-f\left( 3 \right) \right]+\left[ f\left( 3 \right)-f\left( 4 \right) \right]+.........+\left[ f\left( n-1 \right)-f\left( n \right) \right]+\left[ f\left( n \right)-f\left( n+1 \right) \right] \\
 & \Rightarrow \sum\limits_{r=1}^{n}{{{T}_{r}}}=f\left( 1 \right)-f\left( n+1 \right) \\
\end{align}\]
Now, as per our assumption $f\left( r \right)=\dfrac{2r+3}{2r\left( r+1 \right)}$ . Then,
\[\begin{align}
  & \sum\limits_{r=1}^{n}{{{T}_{r}}}=f\left( 1 \right)-f\left( n+1 \right) \\
 & \Rightarrow \sum\limits_{r=1}^{n}{{{T}_{r}}}=\dfrac{2+3}{2\times 1\times 2}-\dfrac{2n+2+3}{2\left( n+1 \right)\left( n+2 \right)} \\
 & \Rightarrow \sum\limits_{r=1}^{n}{{{T}_{r}}}=\dfrac{5}{4}-\dfrac{2n+5}{2\left( {{n}^{2}}+3n+2 \right)} \\
\end{align}\]
Now from the above result, we conclude that the sum of the first $n$ terms of the given series will be \[\sum\limits_{r=1}^{n}{{{T}_{r}}}=\dfrac{5}{4}-\dfrac{2n+5}{2\left( {{n}^{2}}+3n+2 \right)}\] .
Now, the sum of infinite terms will be \[\sum\limits_{r=1}^{\infty }{{{T}_{r}}}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{5}{4}-\dfrac{2n+5}{2\left( {{n}^{2}}+3n+2 \right)} \right]\] . Then,
\[\begin{align}
  & \sum\limits_{r=1}^{\infty }{{{T}_{r}}}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{5}{4}-\dfrac{2n+5}{2\left( {{n}^{2}}+3n+2 \right)} \right] \\
 & \Rightarrow \sum\limits_{r=1}^{\infty }{{{T}_{r}}}=\underset{n\to \infty }{\mathop{\lim }}\,\left[ \dfrac{5}{4}-\dfrac{2+{}^{5}/{}_{n}}{2\left( n+3+{}^{2}/{}_{{{n}^{2}}} \right)} \right] \\
 & \Rightarrow \sum\limits_{r=1}^{\infty }{{{T}_{r}}}=\left[ \dfrac{5}{4}-0 \right] \\
 & \Rightarrow \sum\limits_{r=1}^{\infty }{{{T}_{r}}}=\dfrac{5}{4} \\
\end{align}\]
Now, from the above result, we can say that sum of infinite terms of the given series will be \[\sum\limits_{r=1}^{\infty }{{{T}_{r}}}=\dfrac{5}{4}\] .
Thus, final answers will be \[\sum\limits_{r=1}^{n}{{{T}_{r}}}=\dfrac{5}{4}-\dfrac{2n+5}{2\left( {{n}^{2}}+3n+2 \right)}\] and \[\sum\limits_{r=1}^{\infty }{{{T}_{r}}}=\dfrac{5}{4}\] where ${{T}_{r}}$ is the ${{r}^{th}}$ term of the given series.

Note: Here, the student should first try to understand what is asked in the question. After that, we should first try to analyse the given series and somehow write the expression of ${{r}^{th}}$ term. We should write it correctly without any mistake and then write in the form of ${{T}_{r}}=f\left( r \right)-f\left( r+1 \right)$ so, that we will be able to calculate the value of summation easily without any mistake.
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