Finite sum which sums to $x e^x$ In my analysis class, we have to prove this and another two related sums that I think I could prove if I knew this. However, I don't know how to begin solving this; any hints would be appreciated.
$$\sum_{n=0}^{\infty} \left( e^x-1-\frac{x}{1!}-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots-\frac{x^n}{n!} \right)=xe^x$$
One method of proving this mentions summation by parts, and the only reference I found in relation to that is Abel's lemma which I tried to use but doesn't help me. What other solutions are there?
 A: Let's write the sum in question using summation notation:
\begin{align*}
S&=\sum_{n=0}^\infty \left(e^x-\sum_{k=0}^n\frac{x^k}{k!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \frac{x^k}{k!}-\sum_{k=0}^n \frac{x^k}{k!}\right)\\
&=\sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!}
\end{align*}
Let's switch the order of summation:
$$S=\sum_{k=1}^\infty\sum_{n=0}^{k-1}\frac{x^k}{k!}$$
But note: the value of the summand doesn't depend on $n$. Therefore, we can treat that inner sum like we have a constant summand, in which case we're just multiplying by $k=(k-1)-0+1$. So,
$$S=\sum_{k=1}^\infty k\cdot\frac{x^k}{k!}=\sum_{k=1}^\infty\frac{x^k}{(k-1)!}$$
Let $k=m+1$:
$$S=\sum_{m=0}^\infty \frac{x^{m+1}}{m!}=x\sum_{m=0}^\infty\frac{x^m}{m!}=xe^x$$
as required.
A: \begin{align}
& \sum_{n=0}^\infty \left( e^x-1-\frac x {1!}-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots-\frac{x^n}{n!} \right) = \sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!} \\[8pt]
& \begin{array}{cccccccccc}
= & x & + & \dfrac{x^2} 2 & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & + & \dfrac{x^2} 2 & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & + & \dfrac{x^3} 6 & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & & & + & \dfrac{x^4}{24} & + & \dfrac{x^5}{120} & + & \cdots \\[8pt]
& & & & & & & & + & \dfrac{x^5}{120} & + & \cdots
\end{array} \\[10pt]
= {} & x + x^2 + \frac{x^3} 2 + \frac{x^4} 6 + \frac{x^5}{24} + \cdots \\[8pt]
= {} & x\left( 1 + x + \frac{x^2} 2 + \frac{x^3} 6 + \frac{x^4}{24} +\cdots \right) = xe^x.
\end{align}
Postscript: Maybe it is useful to express it in a way in which the general form is explicit rather than being indicated by three dots that mean "continue with the same pattern."
\begin{align}
& \sum_{n=0}^\infty \sum_{k=n+1}^\infty \frac{x^k}{k!} \\[8pt]
= {} & \sum_{n,k\,:\,k\,\ge\,n+1\,\ge\,1} \frac{x^k}{k!} \\[8pt]
= & \sum_{k=1}^\infty \left( \sum_{n=0}^{k-1} \frac{x^k}{k!} \right)
\end{align}
But no $\text{“}n\text{”}$ appears in this sum in which $n$ goes from $0$ to $k-1$; therefore the sum is just $x^k/k!$ multiplied by the number of terms, which is $k$:
$$
\sum_{k=1}^\infty k\cdot\frac{x^k}{k!} = x\sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!} = xe^x.
$$
A: Hint :
$$\sum_{n=0}^{+\infty}  \sum_{k={n+1}}^{+\infty} \frac{x^k}{k!}=\sum_{k=1}^{+\infty}  \sum_{n=0}^{k-1} \frac{x^k}{k!} = \sum_{k=1}^{+\infty}  k\times\frac{x^k}{k!} = \sum_{k=0}^{+\infty}  \frac{x^{k+1}}{k!} = x e^x$$
A: We can start with the well known Maclaurin series for
$e^x$
