Question

The value of $\dfrac{5}{{1 \cdot 2 \cdot 3}} + \dfrac{7}{{3 \cdot 4 \cdot 5}} + \dfrac{9}{{5 \cdot 6 \cdot 7}} + ...$
A. $\log \left( {\dfrac{8}{e}} \right)$
B. $\log (8e)$
C. $\log \left( {\dfrac{e}{8}} \right)$
D. None of these

Answer 1

Hint: Find the general term of the given series and then use the rule of partial sum fraction. After that use the expansion of $\log (1 + x)$.

Formula Used:
$n^{th}$-term of an A.P. is $a + \left( {n - 1} \right)d$, where a is the first term and d is the common difference of the A.P.
The expansion of $\log (1 + x)$ is $x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ...$

Complete step by step solution:
Let the sum of the given series be S.
Then
$S=\frac{5}{1\cdot \:2\:\cdot \:3}+\frac{7}{3\cdot \:4\cdot \:5}+\frac{9}{5\cdot \:6\cdot \:7}$
We have to find the general term i.e. the n-th term of the series.
Clearly, the numerators of the given fractions are in A.P.
Find n-th term of the A.P. $5,7,9,...$
The first term of this A.P. is 5 and the common difference is $7 - 5 = 9 - 7 = 2$
So, n-th term is $5 + \left( {n - 1} \right) \times 2 = 5 + 2n - 2 = 2n + 3$
The denominators of the given fractions are $\left( {1 \cdot 2 \cdot 3} \right),\left( {3 \cdot 4 \cdot 5} \right),\left( {5 \cdot 6 \cdot 7} \right),...$
The first components $1,3,5,...$ are in A.P., where the first term is 1 and the common difference i$3 - 1 = 5 - 3 = 2$
So, the n-th term is $1 + \left( {n - 1} \right) \times 2 = 1 + 2n - 2 = 2n - 1$
The second components $2,4,6,...$ are in A.P., where the first term is 2 and the common difference is $4 - 2 = 6 - 4 = 2$
So, the n-th term is $2 + \left( {n - 1} \right) \times 2 = 2 + 2n - 2 = 2n$
The third components $3,5,7,...$ are in A.P., where the first term is 3 and the common difference is $5 - 3 = 7 - 5 = 2$
So, the n-th term is $3 + \left( {n - 1} \right) \times 2 = 3 + 2n - 2 = 2n + 1$
Finally, we get the n-th term i.e. the general term of the given series as $\dfrac{{2n + 3}}{{\left( {2n - 1} \right)\left( {2n} \right)\left( {2n + 1} \right)}}$
We have to find the value of the infinite series$S = \sum\limits_{n = 1}^\infty {\dfrac{{2n + 3}}{{(2n - 1)(2n)(2n + 1)}}} ......(i)$
Let $\dfrac{{2n + 3}}{{\left( {2n - 1} \right)\left( {2n} \right)\left( {2n + 1} \right)}} = \dfrac{A}{{2n - 1}} + \dfrac{B}{{2n}} + \dfrac{C}{{2n + 1}}.....(ii)$
Then
$\dfrac{{2n + 3}}{{\left( {2n - 1} \right)\left( {2n} \right)\left( {2n + 1} \right)}} = \dfrac{{A(2n)(2n + 1) + B(2n - 1)(2n + 1) + C(2n - 1)(2n)}}{{(2n - 1)(2n)(2n + 1)}}\\ \Rightarrow 2n + 3 = A(4{n^{^2}} + 2n) + B(4{n^2} - 1) + C(4{n^2} - 2n)\\ \Rightarrow 2n + 3 = 4A{n^2} + 2An + 4B{n^2} - B + 4C{n^2} - 2Cn\\ \Rightarrow 2n + 3 = 4(A + B + C){n^2} + 2(A - C)n - B$
Equating the coefficients of like terms, we get
$A + B + C = 0.....(iii)\\2(A - C) = 2.....(iv)\\ - B = 3.....(v)$
From $(v)$, we get $B = - 3$
From $(iv)$, we get $A - C = 1 \Rightarrow A = 1 + C.....(vi)$
Putting the value of B and substituting $A = 1 + C$ in $(iii)$, we get
$1 + C - 3 + C = 0\\ \Rightarrow - 2 + 2C = 0\\ \Rightarrow 2C = 2\\ \Rightarrow C = 1$
Putting the value of $C$ in $(vi)$, we get A=1+1=2
Now, we have A=2, B=-3, C=1
Putting the values of A, B and C in equation i, we get
$\dfrac{{2n + 3}}{{(2n - 1)(2n)(2n + 1)}} = \dfrac{2}{{2n - 1}} - \dfrac{3}{{2n}} + \dfrac{1}{{2n + 1}}$
$ = \dfrac{2}{{2n - 1}} - \dfrac{2}{{2n}} - \dfrac{1}{{2n}} + \dfrac{1}{{2n + 1}}\\ \Rightarrow 2\left( {\dfrac{1}{{2n - 1}} - \dfrac{1}{{2n}}} \right) - \left( {\dfrac{1}{{2n}} - \dfrac{1}{{2n + 1}}} \right)$
Now, substitute the above expression in $(i)$, we get
$S = \sum\limits_{n = 1}^\infty {\left[ {2\left( {\dfrac{1}{{2n - 1}} - \dfrac{1}{{2n}}} \right) - \left( {\dfrac{1}{{2n}} - \dfrac{1}{{2n + 1}}} \right)} \right]} $
Now, we can write $\sum\limits_n {\left\{ {a{f_n}\left( x \right) + b{g_n}\left( x \right)} \right\} = a\sum\limits_n {{f_n}\left( x \right)} + b\sum\limits_n {{g_n}\left( x \right)} } $
So, $S = 2\sum\limits_{n = 1}^\infty {\left( {\dfrac{1}{{2n - 1}} - \dfrac{1}{{2n}}} \right) - \sum\limits_{n = 1}^\infty {\left( {\dfrac{1}{{2n}} - \dfrac{1}{{2n + 1}}} \right)} } $
Put $n = 1,2,3,......$ successively
$S = 2\left\{ {\left( {1 - \dfrac{1}{2}} \right) + \left( {\dfrac{1}{3} - \dfrac{1}{4}} \right) + \left( {\dfrac{1}{5} - \dfrac{1}{6}} \right) + ......} \right\} - \left\{ {\left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) + \left( {\dfrac{1}{4} - \dfrac{1}{5}} \right) + \left( {\dfrac{1}{6} - \dfrac{1}{7}} \right) + ......} \right\}.....(vii)$
The expansion of $\log (1 + x)$ is $x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + ...$
Put $x = 1$
$\log (2) = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + .....\\ \Rightarrow 1 - \log (2) = \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{4} - .....$
Substitute the above expansions in the series $(vii)$
$S = 2\log (2) - (1 - \log 2)\\ \Rightarrow \log 4 - 1 + \log 2\\ \Rightarrow \log 4 + \log 2 - 1\\ \Rightarrow \log 8 - \log e\\ \Rightarrow \log \left( {\dfrac{8}{e}} \right)$

Option ‘A’ is correct

Note: In this problem, the numerators and the components of the denominators are in A.P. because the difference between the terms are equal. Instead of A.P. it may be given as G.P. or H.P. For G.P., the ratios of the successive terms must be equal and for H.P. the differences of the reciprocals of the successive terms must be equal. You should be aware of the expansions of some special functions.