Question

$\log_{e}2 + \log_{e}\left( 1 + \frac{1}{2} \right) + \log_{e}\left( 1 + \frac{1}{3} \right) + .... + \log_{e}\left( 1 + \frac{1}{n - 1} \right)$ is

equal to.

• A. $\log_{e}1$
• B. $\log_{e}n$
• C. $\log_{e}(1 + n)$
• D. $\log_{e}(1 - n)$

Answer 1

Answer: (B)

$\log_{e}n$

Answer 2

The given series reduces to

$\frac{e^{4} - 1}{4e^{2}}$

$\frac{e^{4} + 1}{4e^{2}}$....

$\frac{1^{2}.2}{1!} + \frac{2^{2}.3}{2!} + \frac{3^{2}.4}{3!} + .....\infty = 6e$.