Because I needed to evaluate the series $$S=\displaystyle\sum_{k=1}^{+\infty}\dfrac{1}{2^k+k!}$$ using the Milne inequality, I found for it, the bound: $$S\lt1-\dfrac{1}{e}.$$ Is it possible to have a sharper bound for S? Thanks
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asked May 18, 2016 at 14:51
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$\begingroup$ The actual value is about 0.604504, by the way. $\endgroup$– Patrick StevensMay 18, 2016 at 14:55
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$\begingroup$ That series is rapidly convergent anyway. I suppose you're not interested in accelerations like $$\sum_{k=1}^m\left(\frac1{2^k+k!}-\frac1{k!}+\frac{2^k}{(k!)^2}\right)+e-I_0(2 \sqrt{2})+\sum_{k=m+1}^{\infty}\frac{2^{2k}}{(k!)^2(2^k+k!)}$$ $\endgroup$– user5713492May 18, 2016 at 22:18
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2 Answers
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I have a silly way to bound it by $\frac{1300811445957}{2144935815680}$:
$$\sum_{k=10}^{\infty} \frac{1}{2^k} + \sum_{k=1}^9 \frac{1}{2^k + k!}$$
answered May 18, 2016 at 14:59
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$\begingroup$ Is this sharper? I don't want use calculator;-) $\endgroup$ May 18, 2016 at 15:02
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$\begingroup$ Yeah, it's quite a lot sharper. It's about 0.606457, as opposed to $1-e^{-1} = 0.632121$. $\endgroup$ May 18, 2016 at 15:05
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$$\begin{eqnarray*} S = \sum_{k=1}^{4}\frac{1}{2^k+k!}+\sum_{k\geq 5}\frac{1}{2^k+k!}&\stackrel{AM-GM}{\leq}&\frac{167}{280}+\frac{1}{2}\sum_{k\geq 5}\frac{1}{2^{k/2}\sqrt{k!}}\\&\stackrel{CS}{\leq}&\frac{167}{280}+\frac{1}{2}\sqrt{\sum_{k\geq 5}\frac{1}{2^k}\sum_{k\geq 5}\frac{1}{k!}} \end{eqnarray*}$$ gives the sharper upper bound: $$ S \leq \frac{167}{280}+\frac{1}{8}\sqrt{\frac{24e-65}{24}}\leq \color{red}{0.6089}. $$ It becomes even sharper (but messier) if you replace the threshold $k=4$ above by something bigger.
answered May 18, 2016 at 15:25