Question

\[\dfrac{1}{{1.3}} + \dfrac{1}{2}.\dfrac{1}{{3.5}} + \dfrac{1}{3}.\dfrac{1}{{5.7}} + ......\infty = \]
A) \[2{\log _e}2 - 1\]
B) \[{\log _e}2 - 1\]
C) \[{\log _e}2\]
D) None of these

Answer 1

Hint: in this question, we have to find the sum of the given infinite series. In order to solve this first Rearrange the given series to find standard pattern of the series. Once we get type of series then by applying the formula of that series, the required sum is to be calculated.



Formula Used:Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is
\[{\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \dfrac{{{x^5}}}{5} + ...\]
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is
\[{\log _e}(1 - x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} - \dfrac{{{x^5}}}{5} - ...\]



Complete step by step solution:Given: \[\dfrac{1}{{1.3}} + \dfrac{1}{2}.\dfrac{1}{{3.5}} + \dfrac{1}{3}.\dfrac{1}{{5.7}} + ......\infty \]
Now we have to rearrange the above series in order to find standard pattern of series.
\[\dfrac{1}{{1.3}} + \dfrac{1}{2}.\dfrac{1}{{3.5}} + \dfrac{1}{3}.\dfrac{1}{{5.7}} + ......\infty = \dfrac{1}{2}(\dfrac{1}{1} - \dfrac{1}{3}) + \dfrac{1}{{{2^2}}}(\dfrac{1}{3} - \dfrac{1}{5}) + ...\]
\[ = (\dfrac{1}{{1.2}} - \dfrac{1}{{2.3}}) + (\dfrac{1}{{3.4}} - \dfrac{1}{{4.5}}) + ...\]
\[\dfrac{1}{{1.2}} - \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} - \dfrac{1}{{4.5}} + ...........\infty = (1 - \dfrac{1}{2}) - (\dfrac{1}{2} - \dfrac{1}{3}) + (\dfrac{1}{3} - \dfrac{1}{4}) - (\dfrac{1}{4} - \dfrac{1}{5}) + ...\]
\[(1 - \dfrac{1}{2}) - (\dfrac{1}{2} - \dfrac{1}{3}) + (\dfrac{1}{3} - \dfrac{1}{4}) - (\dfrac{1}{4} - \dfrac{1}{5}) + ... = (1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5}) + ( - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5})\]……………………… (i)
We know that
Expansion of logarithm series is given as:
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is
\[{\log _e}(1 + x) = x - \dfrac{{{x^2}}}{2} + \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} + \dfrac{{{x^5}}}{5} + ...\]
Put \[x = 1\]
Now we get
\[\log (2) = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6}....\]
We also know that
If\[x \in R\], \[\left| x \right| < 1\], then the expansion is
\[{\log _e}(1 - x) = - x - \dfrac{{{x^2}}}{2} - \dfrac{{{x^3}}}{3} - \dfrac{{{x^4}}}{4} - \dfrac{{{x^5}}}{5} - ...\]
Put \[x = - 1\]
Now we get
\[\log (2) = 1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6}....\] (ii)
\[\log (2) - 1 = - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5} - \dfrac{1}{6}....\] (iii)
Now use equation (ii) and equation (iii) in equation (i) we get
\[(1 - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5}) + ( - \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{5}) = \log (2) + \log (2) - 1\]
\[ = 2{\log _e}2 - 1\]



Option ‘A’ is correct



Note: Here we have to rearrange the series in order to get standard pattern of series. Once we get type of series then by applying the formula of that series, the required sum is to be calculated.
In this question after rearrangement we found that series is written as a sum of expansion of\[{\log _e}(1 + x)\]After getting standard series we have to apply the expansion formula to get the sum of given series. In this type of question always try to find the pattern of the series and after getting pattern apply formula of that pattern.