Question

$3 + \frac{5}{1!} + \frac{7}{2!} + \frac{9}{3!} + .....\infty =$

• A. $3e$
• B. $5e$
• C. $5e - 1$
• D. None of these

Answer 1

Answer: (B)

$5e$

Answer 2

$\frac{4e + 1}{6}$

$\frac{2}{1!}\log_{e}2 + \frac{2^{2}}{2!}(\log_{e}2)^{2} + \frac{2^{3}}{3!}(\log_{e}2)^{3} + .....\infty = 1 + \frac{\log_{e}x}{1!} + \frac{(\log_{e}x)^{2}}{2!} + \frac{(\log_{e}x)^{3}}{3!} + .....\infty =$

Now, sum$\log_{e}x$

$x$.