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I would like to find an equivalent of $\sum_{k=n}^\infty \frac{1}{k!}$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.

$$n!\sum_{k=n}^\infty \frac{1}{k!}=1+\frac{1}{n+1}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots$$ $$=1+\frac{1}{n+1}+\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right)=1+o(1).$$ and thus the claim follow.

Q1) Is it correct ?

Q2)

I have that $\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right)$ because $$\frac{1}{(n+1)(n+2)}\leq \frac{1}{(n+1)^2},\quad \frac{1}{(n+1)(n+2)(n+3)}\leq \frac{1}{(n+2)^2},\quad \frac{1}{(n+1)(n+2)(n+3)(n+4)}\leq\frac{1}{(n+3)^2} \cdots$$

and so $\frac{1}{(n+1)(n+2)\cdots(n+k)}\leq \frac{1}{(n+k)^2}$. But suppose we could only prove that $$\frac{1}{(n+1)(n+2)\cdots(n+k)}\leq \frac{C_k}{(n+k)^2},$$ i.e. that the constant depend on $k$ and we can't have a uniform bound. Could I conclude that $$ \frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+\cdots=\mathcal O\left(\sum_{k=n+1}^\infty \frac{1}{k^2}\right) \ \ ?$$

Michael Hardy
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Peter
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1 Answers1

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See this post: Calculate $\lim_{n \rightarrow \infty}$ ($n!e-[n!e]$)?

It follows that the integer part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=1}^{n-1} \frac{1}{k!}$, so the fractional part of $(n-1)!e$ is $(n-1)!\sum\limits_{k=n}^{\infty} \frac{1}{k!}$. Hence, an equivalent expression is $\frac{\{(n-1)!e\}}{(n-1)!}$.

A. Pongrácz
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  • Sorry but it doesn't answer my question. I know it's equivalent. I just want to know if my method is correct, and if my Q2) is right. – Peter Jul 31 '18 at 07:25
  • Actually, it does. If you check the post I linked, the same calcualtion is also carried out there, showing that your calculation is correct. Your question was not specific. You answer was that $\frac{1}{n!}$ is asymptotically equal to the expression, mine was that $\frac{{(n-1)!e}}{(n-1)!}$ is EQUAL to the expression. As the question was not concrete, anyone can pick their favourite answer, and I prefer mine to yours. With that being said, yours was also correct. – A. Pongrácz Jul 31 '18 at 09:56