$(1 + 3)\log_e 3 + \frac{1 + 3^2}{2!} (\log_e 3)^2 + \frac{1 + 3^3}{3!} (\log_e 3)^3 + \dots \infty = $

$b = 1 + \frac{{}^1 C_0 + {}^1 C_1}{1!} + \frac{{}^2 C_0 + {}^2 C_1 + {}^2 C_2}{2!} + \frac{{}^3 C_0 + {}^3 C_1 + {}^3 C_2 + {}^3 C_3}{3!} + \ldots$
Let $a = 1 + \frac{{}^2 C_2}{3!} + \frac{{}^3 C_2}{4!} + \frac{{}^4 C_2}{5!} + \ldots$. Then $\frac{2b}{a^2}$ is equal to:

The sum of the series $\frac{1}{1 \times 2} + \frac{1 \times 3}{1 \times 2 \times 3 \times 4} + \frac{1 \times 3 \times 5}{1 \times 2 \times 3 \times 4 \times 5 \times 6} + \dots \infty$ is

In the expansion of $\frac{1 - 2x + 3x^2}{e^x}$,the coefficient of $x^5$ will be

If $x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$ and $y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$, then the value of $\log_{e} y$ is

Find the sum of the series $\frac{1}{2!} + \frac{1}{4!} + \frac{1}{6!} + \dots \infty$.