Prove that $\sum_{k=0}^{n}{\frac{1}{k!}}\leq 3$ $\forall n\in \mathbb{N}$
I proved it by induction:
"$n=1$" $ \quad \sum_{k=0}^{1}{\frac{1}{k!}}\leq 3 \quad \checkmark$
"$n \implies n+1$": $$\quad \sum_{k=0}^{n+1}{\frac{1}{k!}}\leq 3 \\ \left(\sum_{k=0}^{n}{\frac{1}{k!}}\right)+\frac{1}{(n+1)!}\leq 3 \\ $$$$3+\overbrace{\frac{1}{(n+1)!}}^{\geq 0}\leq 3 \\$$ That's of course true.
On
For any finite $n$,
$$\sum_{k=0}^n \frac{1}{k!} < e < 3.$$
UPDATE
$$1+1+\frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n! }< 1 + 1+ \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^{n-1}} = 1+2 - \frac{1}{2^{n-1}} <3.$$
On
That's not a correct proof.
Just because $x<3$ does not mean $x+\frac{1}{(n+1)!}<3$ as well. For instance, if $x=2.999999$ and $n=4$ this fails.
To solve this question, one way would be to note that for $n>1,\frac{1}{n!}\le\frac{1}{n(n-1)}$ so we have
$$\sum_{i=0}^\infty \frac{1}{i!}=1+1+\sum_{i=2}^\infty\frac{1}{i!}\le2+\sum_{i=2}^\infty\frac{1}{i(i-1)}=2+\sum_{i=2}^\infty\frac{1}{i-1}-\frac{1}{i}=3$$
where the last equality comes from the fact that we have a telescoping sum of the form $$\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots$$
On
For each $k$ we have $${1\over (k+1)!}<{1\over 2k!}$$
So
$$\sum_{k=0}^{n+1}{\frac{1}{k!}}=1+ 1+{1\over 2!} +{1\over 3!}+...+{1\over (n+1)!}$$ $$<2+{1\over 2\cdot 1!} +{1\over 2\cdot 2!}+...+{1\over 2\cdot n!}$$ $$=2+{1\over 2}\Big(\underbrace{1+{1\over 2!} +{1\over 3!}+...+{1\over n!}}_{<3-1}\Big)$$ $$<2+{3-1\over 2} <3$$
On
$$\sum_{k=0}^{\infty}\frac{1}{k!} = 1 + 1 + \frac{1}{2!}+\frac{1}{3!}+ \cdots$$ $$ < 1+1+\frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4} +\frac{1}{4\cdot 5}+ \cdots$$ $$ = 1 + 1 +\left (1-\frac{1}{2}\right) + \left (\frac{1}{2}-\frac{1}{3}\right) + \left (\frac{1}{3}-\frac{1}{4}\right)+\cdots$$ $$ = 1+1 +1 =3.$$
\begin{align*} \sum_{k=0}^n\frac1{k!}&=2+\sum_{k=2}^n\frac1{k!}\\ &=2+\sum_{k=2}^n\prod_{i=2}^k\frac1i\\ &\le2+\sum_{k=2}^n\prod_{i=2}^k\frac12\\ &=2+\sum_{k=2}^n\frac1{2^{k-1}}\\ &<2+\sum_{k=2}^\infty\frac1{2^{k-1}}\\ &=2+\frac{\frac12}{1-\frac12}\\ &=3 \end{align*}