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If $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.

The base case where $n=1$ is clearly true: $\frac{1}{2!}=\frac{1}{2}$ and $1-\frac{1}{(1+1)!}=\frac{1}{2}$.

Now to show that $S_k\Rightarrow S_{k+1}$, we assume that $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k}{(k+1)!}=1-\frac{1}{(k+1)!}$ and observe the following equalitites:

$\begin{align*}\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k+1}{((k+1)+1)!}&=\\ (\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{k}{(k+1)!})+\Big(\frac{(k+1)}{((k+1)+1)!}\Big)&=\\ 1-\frac{1}{(k+1)!}+\frac{k+1}{((k+1)+1)!}&=\frac{(k+2)(k+1)!-(k+2)+(k+1)}{(k+2)(k+1)!}\\ &=\frac{(k+2)(k+1)!-1}{(k+2)(k+1)!}\\ &=1-\frac{1}{(k+2)!}\\ &=1-\frac{1}{((k+1)+1)!}\end{align*}$

Therefore we conclude that if $n\in\Bbb N$ then $\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$.

I know this is a relatively straightforward example but I'm very new to the concept of induction and I'm wondering if I'm on the right track with this proof. It seems correct to me but I'm worried I may have overlooked something. Any advice is helpful, thanks!

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    $\begingroup$ With $\dfrac{n}{(n+1)!}=\dfrac {n+1}{(n+1)!}-\dfrac1{(n+1)!}=\dfrac1{n!}-\dfrac1{(n+1)!}$, it becomes a telescoping sum $\endgroup$ Dec 18, 2019 at 18:28
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    $\begingroup$ If you add $\frac{1}{(n+1)!}$ to both sides, the inductive step becomes really obvious. $\endgroup$
    – saulspatz
    Dec 18, 2019 at 18:56

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Your proof is correct. Very well done. However, I would strongly recommend familiarising yourself with using sums as an easier way to express your work.

We are proving:

$$\sum_{r=1}^n\frac{r}{(r+1)!}=1-\frac{1}{(n+1)!} \tag A$$

1) Your base case was done well.

2) Assumption step, assume true for $n=k$ $$\sum_{r=1}^k\frac{r}{(r+1)!}=1-\frac{1}{(k+1)!}\tag B $$

3) Use to show for $n=k+1$. From $(B)$ we get:

$$\sum_{r=1}^{k+1}\frac{r}{r+1}=1-\frac{1}{(k+1)!}+\frac{k+1}{(k+2)!}$$ While from $(A)$ we get: $$\sum_{r=1}^{k+1}\frac{r}{r+1}=1-\frac{1}{(k+2)!}$$

We show these are identical, which implies:

$$\frac{1}{(k+2)!}\equiv\frac{1}{(k+1)!}-\frac{k+1}{(k+2)!}$$

$$RHS\to \frac{(k+2)!-{(k+1)!}(k+1)}{(k+1)!(k+2)!}\equiv\frac{(k+2)-(k+1)}{(k+2)!}\equiv\frac{1}{(k+2)!} a.r. $$

I use "a.r." for "as required". Just a bit of flair.

You should then conclude in the manner of: "I have shown (statement) to be true for $n=1$ and true for $n=k+1$ when $n=k$ has been assumed, hence it is true for all $n\in\Bbb N$"

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A couple of your steps could possibly be arranged in a more readable way. However, I do believe your proof is core.

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