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Is the following formula already known?

$$\sum_{k=0}^{\infty} \frac{2^k k!}{(2k + 1)!} = \sqrt{\frac{e\pi}{2}} \operatorname{erf}\left(\frac{1}{\sqrt{2}}\right)$$

The right-hand side is known since Ramanujan, but the left-hand side has not been reported in the literature. So I wonder if this connection between these two expressions is already known.

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    $\begingroup$ "The right member is known since Ramanujan but the left member has not been reported in the literature." -- What do you mean by this? In particular, what do you mean by "The right member is known since Ramanujan"? How can a "member" be known or unknown? $\endgroup$
    – Iosif Pinelis
    Commented Apr 9 at 11:30
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    $\begingroup$ @MartinNicholson : Thank you for the references. In some cases, it seems much easier to prove a thing than to find it in the literature. $\endgroup$
    – Iosif Pinelis
    Commented Apr 9 at 17:05
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    $\begingroup$ This was answered in math.stackexchange.com/questions/3063482 and its related linked questions. $\endgroup$
    – Somos
    Commented Apr 9 at 19:02
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    $\begingroup$ @CrawCraw : Do you have a response to the answers below? $\endgroup$
    – Iosif Pinelis
    Commented Apr 11 at 2:12
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    $\begingroup$ It looks like the only actual answer to this question with a pointer to the literature for this identity was in a now-deleted comment of Martin Nicholson. $\endgroup$
    – Sam Hopkins
    Commented Apr 11 at 13:54

2 Answers 2

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Let $$f(x):=\sqrt{\pi } e^{x^2/4} \operatorname{erf}(x/2).$$ Then $$f'(x)=\tfrac12\, x f(x)+1.$$ So, by the Leibniz formula, for $m\ge1$, $$f^{(m+1)}(x)=\tfrac12\,x f^{(m)}(x)+\tfrac m2\,f^{(m-1)}(x)$$ and hence $$f^{(m+1)}(0)=\tfrac m2\,f^{(m-1)}(0).$$ Also, $f(0)=0$ and $f'(0)=1$. So, for all $k\ge0$ we have $f^{(2k)}(0)=0$ and $f^{(2k+1)}(0)=k!$. So, $$\sqrt{\pi } e^{x^2/4} \operatorname{erf}(x/2)=f(x) =\sum_{k=0}^\infty\frac{k!}{(2k+1)!}\,x^{2k+1}. \tag{10}\label{10}$$ Your identity is a special case of \eqref{10} with $x=\sqrt2$.

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  • $\begingroup$ Is it clear that $f(x)$ must equal its Taylor series? $\endgroup$
    – Sam Hopkins
    Commented Apr 9 at 1:14
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    $\begingroup$ @SamHopkins : Yes, because $f$ is an entire function. $\endgroup$
    – Iosif Pinelis
    Commented Apr 9 at 1:16
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    $\begingroup$ If you write \text{erf} instead of \operatorname{erf} then people following your example may type something like $2\text{erf}(x)$ instead of $2\operatorname{erf}(x),$ which looks conspicuously different. Clearly the latter is standard. $\endgroup$
    – Michael Hardy
    Commented Apr 9 at 16:53
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    $\begingroup$ This answer is off topic. $\endgroup$
    – Jannik Pitt
    Commented Apr 9 at 17:08
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    $\begingroup$ @JannikPitt : The post seems to actually contain two questions: the one in the title and the other one being "So I wonder if this connection between these two members already exists." This answer shows that "this connection [...] exists." $\endgroup$
    – Iosif Pinelis
    Commented Apr 9 at 17:13
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WolframAlpha returns this result promptly, so it is probably known or straightforward to show. See here.

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    $\begingroup$ This can be a comment $\endgroup$
    – Тyma Gaidash
    Commented Apr 10 at 12:25

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