What is the value of the following sum?

$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$

The possible answers are:

A. $e$

B. $\frac{e}{2}$

C. $\frac{3e}{2}$

D. $1 + \frac{e}{2}$

I tried to expand the options using the series representation of $e$ and putting in $x=1$, but I couldn't get back the original series. Any ideas?

  • 1
    $\begingroup$ Suggestion - use $n(n+1)/2 = 1+ \cdots + n$. $\endgroup$
    – peter a g
    Jan 2, 2016 at 17:11
  • $\begingroup$ Always check the numerator's degree ! If it is 3 , then you will have to have 4 constants A,B,C,D $\endgroup$ Jan 2, 2016 at 17:34
  • 2
    $\begingroup$ Nicely rendered as $$\frac11+\frac{1+2}{1\times2}+\frac{1+2+3}{1\times2\times3}+\frac{1+2+3+4}{1 \times 2\times3\times4}+\cdots$$ $\endgroup$
    – user65203
    Jan 2, 2016 at 17:49
  • 1
    $\begingroup$ Note that as that this is a multiple-choice question, all but (C) can be quickly eliminated. $\endgroup$
    – Brian Tung
    Jan 2, 2016 at 18:26

6 Answers 6


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

  • 1
    $\begingroup$ How did you know which function to choose $\endgroup$
    – Taylor Ted
    Jan 2, 2016 at 17:26
  • 3
    $\begingroup$ In order to create the extra factors $n+1$ and $n$ in the numerator, I am starting with $\frac{x^{n+1}}{n!}$, therefore we need the series for $x\times e^x$, and we differentiate twice to get the factors. I hope this helps! $\endgroup$ Jan 2, 2016 at 17:38
  • $\begingroup$ When you differentiate twice the series you started with it becomes $\frac{n(n+1)x^{n-1}}{n!}$. What did you put x in here? $\endgroup$
    – Taylor Ted
    Jan 3, 2016 at 2:22


Hence the sum is

$$\frac e2+e.$$

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


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


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 $


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

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

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

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

  • 1
    $\begingroup$ How did you find $U_r$ and why did you write it into partial fractions having specific denominators $\endgroup$
    – Taylor Ted
    Jan 2, 2016 at 17:25
  • 1
    $\begingroup$ Clearly $1+2+3+...+r= \frac{r(r+1)}{2}$ . Thus $ U_r=\frac{r(r+1)}{2r!}$ . Partial fraction is a trick for these kinda Q' s which I learnt in Advanced levels. It works for these type of questions. It is very easy. Nothing to remember $\endgroup$ Jan 2, 2016 at 17:29
  • $\begingroup$ What is AL's ?.Does this trick work for all questions like these(series summing) $\endgroup$
    – Taylor Ted
    Jan 2, 2016 at 17:31
  • $\begingroup$ Yes It works when factorials are there ! $\endgroup$ Jan 2, 2016 at 17:32
  • $\begingroup$ Always check the numerator's degree ! If it is 3 , then you will have to have 4 constants $A,B,C,D$ $\endgroup$ Jan 2, 2016 at 17:34

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*}


Your sum is $$\sum_{n = 1}^{\infty} \sum_{k = 1}^{n} \frac {k} {n!} = \sum_{n = 1}^{\infty} \frac {n + 1} {2 (n - 1)!} = \sum_{n = 0}^{\infty} \frac {n + 2} {2 n!} = \frac {3e} {2}.$$

  • 1
    $\begingroup$ Last equality a little quick, IMO. $\endgroup$
    – user65203
    Jan 2, 2016 at 17:39
  • $\begingroup$ Because $$\sum_{n = 0}^{\infty} \frac {n + 2} {2 n!} = \sum_{n = 0}^{\infty} \frac {1} {2 (n - 1)!} + \frac {1} {n!} = \frac {e} {2} + e.$$ I don't have paper, I don't have pen, I'm sick and in bed, I am eating my sandwich so I'm writing with one hand. $\endgroup$
    – user98186
    Jan 2, 2016 at 17:43

You want calculate $\sum_{k=1}^\infty\frac{k(k+1)}{2(k!)}=\frac{1}{2}\sum_{k=1}^\infty\frac{k+1}{(k-1)!}=\frac{1}{2}\sum_{k=0}^\infty\frac{k+2}{k!}$...(*)

But, as $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$, then $x^2e^x=x^2+x^3+\frac{x^4}{2!}+\frac{x^5}{3!}+\cdots$. Deriving both sides give us $x^2e^x+2xe^x=2x+3x^2+\frac{4x^3}{2!}+\frac{5x^4}{3!}+...$. Thus, evaluating at $x=1$: $e+2e=\sum_{k=0}^\infty\frac{k+2}{k!}$.

Substituying the last in (*) give us finally $\frac{3e}{2}$.


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