How can I prove that 1/1!+1/2!+1/3!...<3? Should I use mathematical induction? I can see that the sum is the same as the power series of e^1 and that is also less than 3. But where is the proof for that?
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This question boils down to proving that $e<3$. And to prove that, you can consult this thread. – Joe Feb 26 '21 at 14:52
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1You are missing a $+1$ at the beginning of the series ? ... & the exercise is probably to show $e<3$ & not assume this. – Donald Splutterwit Feb 26 '21 at 14:53
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Compare it to $\frac{1}{n^2}$ – Ben Feb 26 '21 at 14:54
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1The sum is actually $e-1$. – Peter Feb 26 '21 at 14:55
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$$\sum_{k=1}^\infty\frac1{k!}= 1+\frac12+\frac12\sum_{k=3}^\infty\frac2{k!}\le1+\frac12+\frac12\sum_{k=3}^\infty\frac1{3^{k-2}}=1+\frac12+\frac12\cdot\frac12=\frac74$$
Hagen von Eitzen
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Hint:
For some $m<n$ $$\frac1{1\cdot2\cdot3\cdot4\cdots n}<\frac1{1\cdot2\cdot3\cdot4\cdots m\cdot m\cdot m\cdot m}=\frac1{m!}\frac1{m^{n-m}}.$$
So starting from the $m^{th}$ term, you can bound the series by a geometric one of which you can get the value of the tail. Up to you to find a suitable $m$.