Evaluate \[1+\dfrac{2}{1\cdot 2\cdot 3}+\dfrac{2}{3\cdot 4\cdot 5}+\dfrac{2}{5\cdot 6\cdot 7}+................\infty \]
1. \[2\log 2\]
2. \[2\log 4\]
3. \[2\log 3\]
4. None of these

Hint: First of all we will write the given series and mark it as equation \[(1)\] and then we will take \[2\] common from the series after that we will solve the given series and then we will apply \[\log (1+x)\] formula and put the value of sum of series in equation \[(1)\]to check which option is correct in the given options.

Complete step-by-step solution:
Arya Bhatta was the first to give the formula for the sum of squares and cubes of natural numbers in his book Arya Bhatta. Babylonians the first who know about arithmetic and geometric series. Indian mathematicians, Brahamgupta , Mahavira and Bhaskar contributed very much in this direction.
A sequence is arithmetic if each term \[-\] the previous term = \[d\] where \[d\] is a constant.
When the terms of a sequence are added we get a series and we can derive a formula that can be used for finding the sum of the terms of an arithmetic series.
A list of numbers or objects in a special order is called a sequence.
When you know two terms and the number of terms in a finite arithmetic sequence you can find the sum of the terms.
A series is the indicated sum of the terms of a sequence. A finite series like a finite sequence has a first term and a last term while an infinite series continues without end.
When you add the values in a sequence together, that sum is called a series. Series is of two types : finite series and infinite series.
Now according to the question:
We have given a series:
\[\Rightarrow 1+\dfrac{2}{1\cdot 2\cdot 3}+\dfrac{2}{3\cdot 4\cdot 5}+\dfrac{2}{5\cdot 6\cdot 7}+................\infty \]
Now if we take \[2\] common from the series we will get:
\[\Rightarrow 1+2(\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{3\cdot 4\cdot 5}+\dfrac{1}{5\cdot 6\cdot 7}+................\infty )\] mark it as equation \[(1)\]
Now we will solve for \[\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{3\cdot 4\cdot 5}+\dfrac{1}{5\cdot 6\cdot 7}+................\infty \]
\[\Rightarrow \dfrac{1}{1\cdot 2\cdot 3}=\dfrac{1}{2}(\dfrac{1}{1\cdot 2}-\dfrac{1}{2\cdot 3})\]
\[\Rightarrow \dfrac{1}{3\cdot 4\cdot 5}=\dfrac{1}{2}(\dfrac{1}{3\cdot 4}-\dfrac{1}{4\cdot 5})\]
Now you can split the term \[\dfrac{1}{1\cdot 2}\] as \[1-\dfrac{1}{2}\] and the term \[\dfrac{1}{2\cdot 3}\] as \[\dfrac{1}{2}-\dfrac{1}{3}\]
Hence the sum becomes:
\[\Rightarrow \dfrac{1}{2}(1-\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{4}+.......)\]
\[\Rightarrow \left( \dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6} \right)\]
As we know that
\[\Rightarrow \log (1+x)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-.......\]
Putting the value of \[x=1\] we will get:
\[\Rightarrow \log 2=(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}.......)\]
\[\Rightarrow \log 2-\dfrac{1}{2}\]
Now according to equation \[(1)\]
\[\Rightarrow 1+2(\log 2-\dfrac{1}{2})\]
\[\Rightarrow 1+2\log 2-\dfrac{2}{2}\]
\[\Rightarrow 1+2\log 2-1\]
\[\Rightarrow 2\log 2\]
Hence option \[(1)\] is correct as the sum of series is \[2\log 2\] .

Note: A finite series is given by all the terms of a finite sequence, added together. A series can have a sum only if the individual terms tend to zero. But there are some series with individual terms tending to zero that do not have sums and a sequence of partial sums of a series sometimes tends to a real limit.