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Question:

The sum of the series is:

$\frac{1}{1.2.3}$ + $\frac{1}{3.4.5}$ + $\frac{1}{5.6.7}$ + ....

Explanation:

S = $\sum _{i=0}^{\infty }$$\frac{1}{(2i+1)(2i+2)(2i+3)}$

= $\frac{1}{2}$ $\sum _{i=0}^{\infty }$$\frac{(2i+3)-(2i+1)}{(2i+1)(2i+2)(2i+3)}$

= $\frac{1}{2}$ $\sum _{i=0}^{\infty }$$\left(\frac{1}{(2i+1)(2i+2)}-\frac{1}{(2i+2)(2i+3)}\right)$

= $\frac{1}{2}$$\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+1}-\frac{2}{2i+2}+\frac{1}{2i+3}\right)$

= $\frac{1}{2}$$\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+1}+\frac{1}{2i+3}\right)$ - $\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+2}\right)$            ...(i)

Now, $\frac{1}{2}$$\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+1}+\frac{1}{2i+3}\right)$ = $\frac{1}{2}$$\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{3}+\frac{1}{5}+\frac{1}{5}+\frac{1}{7}+\frac{1}{7}+...\right)$

= $\frac{1}{2}$$\left(1+2\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...\right)\right)$

= $\frac{1}{2}$$\left(2\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...\right)-1\right)$

= $\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...\right)$ - $\frac{1}{2}$

= -$\frac{1}{2}$ + $\sum _{i=0}^{\infty }$ $\left(\frac{1}{2i+1}\right)$                                         ...(ii)

From (i) and (ii), we get,

S = -$\frac{1}{2}$ + $\sum _{i=0}^{\infty }$$\left(\frac{1}{2i+1}-\frac{1}{2i+2}\right)$

= -$\frac{1}{2}$$\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...\right)$

= -$\frac{1}{2}$ + $\sum _{i=0}^{\infty }$${\left(-1\right)}^{(i-1)}$ $\frac{1}{i}$                                ...(iii)

Now, we know that,

log_e (1 + x) = x - $\frac{1}{2}$x² + $\frac{1}{3}$x³ - $\frac{1}{4}$x⁴  + ...

= $\sum _{i=0}^{\infty }$(-1)^{n- 1}  $\frac{1}{n}$xⁿ 

Putting x = 1 in the above equation, we get,

log_e 2 = $\sum _{i=0}^{\infty }$${\left(-1\right)}^{(i-1)}$ $\frac{1}{i}$                            ...(iv)

From (iii) and (iv), we get,

S = log_e 2 - $\frac{1}{2}$

Hence, option (d).

Alternatively,

You can also solve this question using options.

$\frac{1}{1.2.3}$ + $\frac{1}{3.4.5}$ + $\frac{1}{5.6.7}$ + ... = $\frac{1}{6}$ + $\frac{1}{60}$ + $\frac{1}{210}$ + ...

= 0.167 + 0.0167 + 0.0047 + …

< 0.2 but always positive

So, value of S will always be less than 0.2.

Now, we will consider all the given options one by one.

Consider option 1.

e² – 1 = (2.718)² – 1 > 0.2

So, option 1 can be eliminated.

Consider option 2.

log_e2 – 1 = 0.693 – 1 < 0

So, option 2 can be eliminated.

Consider option 3.

2log₁₀2 – 1 = 2(0.3010) – 1 < 0

So, option 3 can be eliminated.

So, the correct answer will be option 4.

Hence, option (d).