Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$

Solution 1:

Hint...start by writing out the series for $xe^x$ and differentiate twice

Solution 2:

$$\frac{k(k+1)}{2k!}=\frac{k(k-1)+2k}{2k!}=\frac1{2(k-2)!}+\frac1{(k-1)!}$$

Hence the sum is

$$\frac e2+e.$$

Note that the first summation in the RHS must be started at $k=2$ (or use $1/(-1)!:=0$).

Solution 3:

Clearly the $r^{th}$ numerator is $1+2+3+...+r= \frac{r(r+1)}{2}$ .

And the $r^{th}$ denominator is $r!$.

Thus $$\displaystyle U_r=\frac{\frac{r(r+1)}{2}}{r!}=\frac{r(r+1)}{2r!}$$

Since the degree of the numerator is $2$ , use partial fractions to find $A,B,C$ such that (If you use partial fractions up to $(r-3)!$ , its' coefficient will be zero when comparing coefficients.)

$\displaystyle U_r=\frac{r(r+1)}{2r!}=\frac{A}{(r-2)!}+\frac{B}{(r-1)!}+\frac{C}{r!}$

$\displaystyle (2r!)\times U_r=(2r!)\times \frac{r(r+1)}{2r!}=(2r!)\times \frac{A}{(r-2)!}+(2r!)\times \frac{B}{(r-1)!}+(2r!)\times \frac{C}{r!}$

So $\displaystyle r(r+1)=r!\times \frac{2A}{(r-2)!}+r!\times \frac{2B}{(r-1)!}+r!\times \frac{2C}{r!}$

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Now observe that

$r!=1\times 2\times 3\times .... \times (r-2)\times(r-1)\times r $

$\Rightarrow r!=(r−2)! ×(r−1)r $ and

$ \Rightarrow r!=(r−1)!×r $

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So $\displaystyle r(r+1)=(r−2)! ×(r−1)r \times \frac{2A}{(r-2)!}+(r−1)!×r\times \frac{2B}{(r-1)!}+r!\times \frac{2C}{r!}$

So $\displaystyle r^2+r = 2A(r-1)r+2Br+ 2C $

Clearly $C=0$ , $B=1$ and $A=\frac{1}{2}$

So $\displaystyle U_r=\frac{r(r+1)}{2r!}=\frac{1}{2(r-2)!}+\frac{1}{(r-1)!}$

$\displaystyle \sum_{r=2}^{\infty}U_r= \frac{1}{2} \left( \frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+.....\right)+\left( \frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.....\right)$

$\displaystyle \sum_{r=2}^{\infty}U_r= \frac{1}{2} \left( e\right)+\left( e-1\right)$

$\displaystyle \sum_{r=1}^{\infty}U_r= U_1+\frac{1}{2} \left( e\right)+\left( e-1\right)=1+\frac{e}{2}+e-1 =\frac{3e}{2}$

Solution 4:

Starting with $\sum_{i=1}^ni=\frac{n(n+1)}{2}$, the sum simplifies to \begin{align*} z=\frac{1}{2}\sum_{i=1}^\infty \frac{i(i+1)}{i!}&=\frac{1}{2}\sum_{i=1}^\infty \frac{i+1}{(i-1)!}\\ &=\frac{1}{2}\sum_{i=0}^\infty\frac{i+2}{i!}\\ &=\frac{1}{2}\sum_{i=0}^\infty\left(\frac{i}{i!}+\frac{2}{i!}\right)\\ &=\frac{1}{2}\left(\sum_{i=1}^\infty\frac{1}{(i-1)!}+\sum_{i=0}^\infty\frac{2}{i!}\right)\\ &=\frac{1}{2}\left(\sum_{i=0}^\infty\frac{1}{i!}+2\sum_{i=0}^\infty\frac{1}{i!}\right)\\ &=\frac{1}{2}(e+2e)=\frac{3e}{2} \end{align*}