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?
Let's write the sum in question using summation notation:
\begin{align*}
S&amp;=\sum_{n=0}^\infty \left(e^x-\sum_{k=0}^n\frac{x^k}{k!}\right)\\
&amp;=\sum_{n=0}^\infty \left(\sum_{k=0}^\infty \frac{x^k}{k!}-\sum_{k=0}^n \frac{x^k}{k!}\right)\\
&amp;=\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.
Hint :
$$\sum_{n=0}^{+\infty}  \sum_{k={n+1}}^{+\infty} \frac{x^k}{k!}=\sum_{k=1}^{+\infty}  \sum_{n=0}^{k-1} \frac{x^k}{k!} = \sum_{k=1}^{+\infty}  k\times\frac{x^k}{k!} = \sum_{k=0}^{+\infty}  \frac{x^{k+1}}{k!} = x e^x$$