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.


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 $n\in\Bbb N_{\ge 1}$.

  • $\begingroup$ Okay so I follow how you're saying how $$G(n): 3-\frac{1}{n}$$ and how you've arrange $$P(n+1) = P(n) + \frac{1}{(n+1)!} \leq G(n+1)$$ $$G(n+1) = G(n)-\frac{1}{(n+1)}+\frac{1}{n}$$ - i'm just having trouble why you add the $$\frac{1}{n}$$ to $$3- \frac{1}{n}$$ (G(n)) and how do you get to the bottom part where you show that $$\frac{1}{(n+1)!} \leq \frac{1}{n}-\frac{1}{n+1}$$ $\endgroup$
    – backslash
    Oct 30 '18 at 4:20
  • $\begingroup$ note that $P(n)\le G(n)\Leftrightarrow P(n)-G(n)\le 0$, and you have that $G(n+1)=G(n)-\frac1{n+1}+\frac1n$ (check it using the definition of $G$). Also we have that $P(n+1)=P(n)+\frac1{(n+1)!}$, so $P(n+1)\le G(n+1)$ is the same that $$P(n)+\frac1{(n+1)!}\le G(n)+\frac1n-\frac1{n+1}$$ Now subtracting $G(n)$ to each side of this last inequality we find that it is enough to show that $\frac1{(n+1)!}\le\frac1n-\frac1{(n+1)}$ to prove that $P(n+1)\le G(n+1)$. $\endgroup$
    – Masacroso
    Oct 30 '18 at 4:26
  • $\begingroup$ $G(n+1)=3-\frac1{n+1}=\underbrace{3-\frac1n}_{=\,G(n)}+\frac1n-\frac1{n+1}$ $\endgroup$
    – Masacroso
    Oct 30 '18 at 4:28
  • $\begingroup$ ohhhhhh okay thanks pretty sure i'm seeing what i'm missing - (now obvious) fact that $ P(n) \leq G(n) \Leftrightarrow P(n) - G(n) \leq 0$ and your last comment helping a lot $\endgroup$
    – backslash
    Oct 30 '18 at 4:37
  • 1
    $\begingroup$ Note that $P(n)-G(n)\le 0$ by assumption (by the induction hypothesis) so if we shows that $\frac1{(n+1)!}\le \frac1n-\frac1{n+1}$ it would be also true that $P(n)-G(n)+\frac1{(n+1)!}\le \frac1n-\frac1{n+1}$, what is equivalent to say that $P(n+1)\le G(n+1)$ $\endgroup$
    – Masacroso
    Oct 30 '18 at 6:01

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