<s>[INST] Prove by induction: $\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}$ 
Prove
  $$\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}.$$

My problem with this is that it doesn't hold for the base case: $n=1$. This question is from the book "Abstract Algebra" by Charles Pinter (Page 212, exercise 7). When I go about proving it, I can't reach my goal which is $\frac{(n+1)!-1}{(n+1)!}=1-\frac{1}{(n+1)!}$, for $n=n+1$.
 [/INST] Here's an alternate proof, with generating functions instead of induction.
Note that
$$
\sum_{i=1}^n \frac{i}{(i+1)!}
$$
is the sum of the first $n$ terms of the series
\begin{align*}
\sum_{i=1}^\infty \frac{i x^i}{(i+1)!}
&= x \sum_{i=1}^\infty \frac{i x^{i-1}}{(i+1)!} \\
&= x \frac{d}{dx} \sum_{i=1}^\infty \frac{x^i}{(i+1)!} \\
&= x \frac{d}{dx} \frac{1}{x} \sum_{i=1}^\infty \frac{x^{i+1}}{(i+1)!} \\
&= x \frac{d}{dx} \frac{1}{x} \left[ e^x - x - 1\right] \\
&= x \frac{d}{dx} \left[ \frac{e^x}{x} - 1 - \frac{1}{x} \right] \\
&= x \left[ \frac{xe^x - e^x}{x^2} + \frac{1}{x^2} \right] \\
&= \frac{xe^x - e^x + 1}{x}.
\end{align*}
To get the generating function for the sum of the first $n$ terms,
we then multiply by $\frac{1}{1 - x}$:
\begin{align*}
\frac{xe^x - e^x + 1}{x(1-x)}
&= \frac{(1 - e^x)(1 - x) + x}{x(1-x)} \\
&= \frac{1}{1-x} - \frac{e^x - 1}{x} \\
\end{align*}
The $n$th term of $\frac{1}{1-x}$ is $1$ and the $n$th term of $\frac{e^x - 1}{x}$ is $\frac{1}{(n+1)!}$, so the result is
$$
\boxed{1 - \frac{1}{(n+1)!}}
$$
as desired.
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