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I was constructing a proof through inequalities, but I am having a bit of problem showing the following step: $$\sum_{k=N+1}^\infty \frac {1}{k!} < \frac {1}{N!}$$

Is there any quick way to show this?

Michael Rozenberg
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Relux the Relux
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  • Can you use the power series expansion of the exponential function, and that $2<e<3$ – kingW3 May 13 '17 at 14:31
  • Yes, we can the power series expansion, but we have not shown yet that $2 < e < 3$ – Relux the Relux May 13 '17 at 14:34
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    Hit: the sequence decreases faster than a geometric progression of common ratio $1/(k+1)$. –  May 13 '17 at 14:45
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    See also https://math.stackexchange.com/questions/2221084/c-n-1-frac11-frac12-frac1n-so-prove-e-c-n-le, https://math.stackexchange.com/questions/1557954/how-to-show-sum-k-n-infty-frac1k-leq-frac2n – Martin R May 13 '17 at 17:33

3 Answers3

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$$\sum_{k=N+1}^\infty \frac {1}{k!} =\frac{1}{N!}\left(\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+...\right)<$$ $$<\frac{1}{N!}\left(\frac{1}{N+1}+\frac{1}{(N+1)^2}+...\right)=\frac{1}{N!}\cdot\frac{\frac{1}{N+1}}{1-\frac{1}{N+1}}\leq \frac {1}{N!}$$

Masacroso
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Michael Rozenberg
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    My edit was to correct the last "$<$" to "$\leq$" (e.g. if $N=1$) which does not affect the validity of the proof because of the strict inequality in the first line......+1 – DanielWainfleet May 13 '17 at 17:17
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An alternative proof:

$$\sum_{k=N+1}^\infty\frac1{k!}<\sum_{k=N+1}^\infty\frac{k-1}{k!}=\sum_{k=N+1}^\infty\frac1{(k-1)!}-\sum_{k=N+1}^\infty\frac1{k!}=\frac1{N!}$$

Masacroso
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All we need to prove is :

$$\sum_{k=N+1}^\infty \frac {N!}{k!} < 1$$

Now, since

$$\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+1)(N+2)(N+3)} \ldots <\frac{1}{(N+1)}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+2)(N+3)}$$

Can you see the telescoping sum now, doing that we get

$$\frac{1}{N+1}+\frac{1}{(N+1)(N+2)}+\frac{1}{(N+1)(N+2)(N+3)} \ldots <\frac{2}{N+1} < 1 ~\forall ~N \ge 2$$

Jaideep Khare
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