Algebra - Exponential and logarithmic series

2 Answers

1708

$\hspace{-16}$Given $\bf{1+\frac{1+2}{1!}+\frac{1+2+3}{2!}+\frac{1+2+3+4}{3!}+...........\infty}$\\\\\\ The given series can be written as $\bf{=\frac{1}{2}\sum_{r=1}^{\infty}\frac{r.(r+1)}{(r-1)!}}$\\\\\\ $\bf{=\frac{1}{2}\sum_{r=1}^{\infty}\frac{r^2+r}{(r-1)!} = \frac{1}{2}\sum_{r=1}^{\infty}\frac{(r^2-1)+(r-1)+2}{(r-1)!}}$\\\\\\ $\bf{=\frac{1}{2}\sum_{r=1}^{\infty}\frac{(r+1)}{(r-2)!}+\frac{1}{2}\sum_{r=1}^{\infty}\frac{1}{(r-2)!}+\sum_{r=1}^{\infty}\frac{1}{(r-1)!}}$\\\\\\ $\bf{=\frac{1}{2}\sum_{r=1}^{\infty}\frac{(r-2)+3}{(r-2)!}+\frac{1}{2}\sum_{r=1}^{\infty}\frac{1}{(r-2)!}+\sum_{r=1}^{\infty}\frac{1}{(r-1)!}}$\\\\\\ $\bf{=\frac{1}{2}\sum_{r=1}^{\infty}\frac{1}{(r-3)!}+\frac{3}{2}\sum_{r=1}^{\infty}\frac{1}{(r-2)!}+\frac{1}{2}\sum_{r=1}^{\infty}\frac{1}{(r-2)!}+\sum_{r=1}^{\infty}\frac{1}{(r-1)!}}$\\\\\\ Now using the formula.....\\\\\\ $\bf{\sum_{r=0}^{\infty}\frac{1}{r!}=e}$\\\\\\ So sum $\bf{=\frac{1}{2}e+\frac{3}{2}e+\frac{1}{2}e+e = 2e+\frac{3}{2}e = \frac{7}{2}e}$$