Have had a lot of trouble with this question. Cannot seem to use the components from the induction hypothesis properly.
Prove by induction:
P(n): $1+\frac{1}{1!}+\frac{1}{2!}+…+\frac{1}{n!} \leq 3 – \frac{1}{n}$ for all $n \in \Bbb N$
So i'm able to prove the base case:
P(1): $1+\frac{1}{(1)!} \leq 3 – \frac{1}{1} = 2 \leq 2$ is true.
Induction:
Assume that
P(k): $1+\frac{1}{1!}+\frac{1}{2!}+…+\frac{1}{k!} \leq 3 – \frac{1}{k}$
is true for some $k \in \mathbb N$
Have to show that
P(k+1): $1+\frac{1}{1!}+\frac{1}{2!}+…+\frac{1}{k!}+\frac{1}{(k+1)!} \leq 3 – \frac{1}{(k+1)}$
So I get up to this step but don't know how to proceed after it:
$1+\frac{1}{1!}+\frac{1}{2!}+…+\frac{1}{k!}+\frac{1}{(k+1)!} \leq 3 – \frac{1}{k}+\frac{1}{(k+1)!}$
(using the $3-\frac{1}{k}$ from the induction hypothesis (P(k)) but I have no idea how to start working with with these factorials to arrive at a conclusion that matches the RHS of P(k+1). I'm thinking there's something to do with that $(k+1)! = (k+1)k!$ but can't connect it all.
Any help would be appreciated – I checked everywhere on here and closest question I found that was similar was this question but it involves equality vs inequality. Thanks hopefully can help. I think this is question is unique enough that it shouldn't be a duplicate.
Best Answer
HINT: let $P(n):=\sum_{k=0}^n\frac1{k!}$ and $G(n):=3-\frac1n$, and we assume by hypothesis that $P(n)\le G(n)$, then you want to show that
$$P(n+1)=P(n)+\frac1{(n+1)!}\le G(n+1)=G(n)-\frac1{n+1}+\frac1n$$
so it is enough to show that $\frac1{(n+1)!}\le\frac1n-\frac1{n+1}$ for all $n\in\Bbb N_{\ge 1}$.