Consider the series: $$ \sum_{i=1}^\infty \frac{i}{(i+1)!} $$ Make a guess for the value of the $n$-th partial sum and use induction to prove that your guess is correct.
I understand the basic principles of induction I think I would have to assume the n-1 sum to be true and then use that to prove that the nth sum is true. But I have no idea how to guess what the sum might be? Doing the partial sums indicates that the series converges at possibly 1.
On
In order to make a guess, you should begin by calculating some partial sums:
$$\begin{align*} \sum_{i=1}^1\frac{i}{(i+1)!}&=\frac12\\\\ \sum_{i=1}^2\frac{i}{(i+1)!}&=\frac12+\frac2{3!}=\frac12+\frac13=\frac56\\\\ \sum_{i=1}^3\frac{i}{(i+1)!}&=\frac56+\frac3{4!}=\frac56+\frac18=\frac{23}{24}\\\\ \sum_{i=1}^4\frac{i}{(i+1)!}&=\frac{23}{24}+\frac4{5!}=\frac{23}{24}+\frac1{30}=\frac{714}{720}=\frac{119}{120} \end{align*}$$
Now look at that sequence of partial sums: $$\dfrac12,\dfrac56,\dfrac{23}{24},\dfrac{119}{120}\;.$$ The pattern of the denominators should leap out at you to suggest a conjecture as to the denominator of the $n$-th partial sum, and the likely relationship between the numerator and the denominator is even more apparent. Write down a conjecture of the form
$$\sum_{i=1}^n\frac{i}{(i+1)!}=\frac{a_n}{b_n}\;,$$
where $a_n$ and $b_n$ are some integer functions of $n$, and try to prove it by induction on $n$. Note that
$$\sum_{i=1}^{n+1}\frac{i}{(i+1)!}=\sum_{i=1}^n\frac{i}{(i+1)!}+\frac{n+1}{(n+2)!}\;,$$
so this boils down to proving that
$$\frac{a_{n+1}}{b_{n+1}}=\frac{a_n}{b_n}+\frac{n+1}{(n+2)!}\;.$$
We have:$$\dfrac{i}{(i+1)!} = \dfrac{(i+1)-1}{(i+1)!} = \dfrac{i+1}{(i+1)!} - \dfrac{1}{(i+1)!} = \dfrac{1}{i!}-\dfrac{1}{(i+1)!} \Rightarrow S_n = 1-\dfrac{1}{(n+1)!}\Rightarrow S=\displaystyle \lim_{n\to \infty} S_n = 1$$