<s>[INST] Representation of $e$ as a descending series I saw on Wikipedia: List of representations of e that  
$$e=3+\sum_{k=2}^{\infty}\frac{-1}{k!(k-1)k}$$ 
It was also mentioned that this identity come from consideration on ways to put upper bound of $e$. 
Can anyone give me a hint how we can derive this identity? I have tried to look at its Taylor expansion but that approach seems to fail miserably. 
Thanks in advance. 
 [/INST] You can derive the following telescoping sum for $k\ge2$:
\begin{align*}
\frac{1}{k!}+\frac{1}{k!(k-1)k}
&=\frac{1}{k!}\left(1+\frac{1}{(k-1)k}\right)\\
&=\frac{1}{k!}\left(1+\frac{1}{k-1}-\frac{1}{k}\right)\\
&=\frac{k-1}{kk!}+\frac{1}{(k-1)k!}\\
&=\frac{k-1}{kk!}+\frac{1}{(k-1)k(k-1)!}\\
&=\frac{k-1}{kk!}+\frac{1}{(k-1)(k-1)!}-\frac{1}{k(k-1)!}\\
&=\frac{1}{k!}-\frac{1}{kk!}+\frac{1}{(k-1)(k-1)!}-\frac{1}{k!}\\
&=\frac{1}{(k-1)(k-1)!}-\frac{1}{kk!}
\end{align*}
Hence, if you go back to the original series representation of $e$ as you did:
\begin{align*}
e+\sum_{k=2}^{\infty}\frac{1}{k!k(k-1)}
&=2+\sum_{k=2}^{\infty}\frac{1}{k!}+\frac{1}{k!k(k-1)}\\
&=2+\sum_{k=2}^{\infty}\frac{1}{(k-1)(k-1)!}-\frac{1}{kk!}\\
&=2+1-\lim_{n\rightarrow\infty}\frac{1}{nn!}=3
\end{align*}
</s>