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Does anyone know a closed form for this expression $$\sum_{i=1}^\infty\frac{c^i}{i^i}$$ where $c$ is an arbitrary constant? It isn't hard to prove it converges for all values of $c$, but I cannot find a formula for this function. I think it might be a Taylor Series of some sort.

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    $\begingroup$ The Wikipedia page on Sophomore's Dream handles the special case where $c=1$, and I'm pretty sure it's not known to have a nice closed form. $\endgroup$ – Carl Schildkraut Sep 12 '16 at 4:06
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    $\begingroup$ Through manipulations similar to those on the link above, one can see that it's equal to $1+\int_0^1 x^{1-cx}\,dx$. Agree with @CarlSchildkraut that it's unlikely to simplify much further. $\endgroup$ – πr8 Sep 12 '16 at 4:17
  • $\begingroup$ @πr8 Are you sure? Wolfram|Alpha disagrees at $c=2$. $\endgroup$ – Carl Schildkraut Sep 12 '16 at 4:46
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    $\begingroup$ Ah, messed up the algebra: the integrand should be $c\cdot x^{-cx}$, accidentally absorbed the wrong power. And because the sum starts at $n=1$ (had previously overlooked), the $+1$ can be ignored $\endgroup$ – πr8 Sep 12 '16 at 8:24
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As already said in comments, closed forms do not exist.

However, consider $$u_n=\frac {c^n}{n^n}$$ and use the ratio test $$\frac{u_{n+1}}{u_n}=c\frac{n^ n}{(n+1)^{n+1}}=\frac cn \left(\frac{n}{n+1}\right)^{n+1}=\frac cn \left(\frac{1}{1+\frac 1n}\right)^{n+1}\to \frac c {en}$$ You could go further writing $$\log\left(\frac{n}c \frac{u_{n+1}}{u_n}\right)=-(n+1)\log\left(1+\frac 1n\right)$$ and use Taylor expansion for large values of $n$; this would give $$\log\left(\frac{n}c \frac{u_{n+1}}{u_n}\right)=-1-\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$ and using $a=e^{\log(a)}$ and Taylor again $$\frac{n}c \frac{u_{n+1}}{u_n}=\frac{1}{e}-\frac{1}{2 e n}+O\left(\frac{1}{n^2}\right)\implies \frac{u_{n+1}}{u_n}=\frac{c}{e n}-\frac{c}{2 e n^2}+O\left(\frac{1}{n^3}\right)$$

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