<s>[INST] How to prove $ \sum\limits_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$ using induction? This is as far as I get.  I get stuck here because both sides to not equal each other, but I am not sure what I am doing wrong.
$$ \sum\limits_{k=1}^{n}\frac{k}{(k+1)!}=1-\frac{1}{(n+1)!}$$
Assume: $ \sum\limits_{k=1}^{c}\frac{k}{(k+1)!} = 1 – 1/(c+1)!$
Prove: $ \sum\limits_{k=1}^{c+1}\frac{k}{(k+1)!} = 1 – 1/((c+1)+1)!$
$$\sum_{k=1}^c \frac{k}{(k+1)!} + \frac{c+1}{((c+1)+1)!} \stackrel{?}{=}  1- \frac{1}{((c+1)+1)!}$$
$$1 – \frac{1}{(c+1)!} + \frac{(c+1)}{((c+1)+1)!} \stackrel{?}{=} 1 – \frac{1}{((c+1)+1)!}$$
 [/INST] Make a common denominator on your left hand side (noting that $(c+1)+1 = c+2$ and that $(c+2)! = (c+2)(c+1)!$):
\begin{eqnarray}
1 - \frac{1}{(c+1)!} + \frac{c+1}{(c+2)!} &=& 1 - \frac{c+2}{(c+1)!(c+2)} + \frac{c+1}{(c+2)!} \\ 
&=&  1 - \frac{c+2}{(c+2)!} + \frac{c+1}{(c+2)!}
\end{eqnarray}
Can you take it from here?
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