$0 \leq C + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} $ ,where $C = 1.036055393...$ is the best possible constant.

Question

How to prove that for all real $x$ we have :

$$0 \leq f(x) = C + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} $$

Where $C = 1.036055393\ldots$ is the best possible constant.

For background see also :

Solve for $m(t)$ in the integral transform $\int_0^1(1-t^n) m(t) dt=\frac{(n+1)^n}{n^{n-1}} $ for $n>0$.

where this originates from.

Also

Conjecture about $ f(x,v) = 1 + \sum_{n=1}^\infty {x^n \over t(v n) } > 0 $ for all real $x$

is within the same spirit.

Related and another possible way to show this might be similar to this here :

Is $ f(x,v) = \sum_{n=0}^\infty {x^n \over \Gamma(v n +1) } > 0 $ for all real $x$ and $0<v<1$?

I assume $C$ has no closed form btw.

edit

Not essential but beyond a comment perhaps :

A similar conjecture

How to prove that for all real $x$ we have :

$$0 \leq f(x) = A + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^{\frac{n}{2}}} $$

Where $A= 1$ is the best possible constant.

edit 2

Notice

$$G(x) = \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} $$

can be analysed with the integral transform from the first link.

$1/(n+1)^n$ at $n = 1$ gives us coefficient $1/2$.

So in order to make use of the integral transform we need

the empty product ($=1$) for the first coefficient.

So we get

$$H(x) = C_H + \sum_{n=1}^{+\infty} \frac{2 x^n}{(n+1)^n} $$

and if

$$\int_0^1(1-t^n) m_h(t) dt=\frac{(n+1)^n}{n^{n-1}} $$ for $n>0$ has a solution with $m_h(t)$ being positive for $0<t<1$

Then

$$H(x) = 1 + \sum_{n=1}^{+\infty} \frac{2 x^n}{(n+1)^n} > 0$$

Is true and therefore

$$H(x)/2 = 1/2 + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} > 0$$

However we know this is NOT TRUE.

$$C = 1.036055393... >> 1/2$$

So the integral transformation can not be done as above.

Nevertheless we seem to get

$$0 \leq f(x) = C + \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} $$

(the subject and question here to solve )

and also

$$ \lim_{x \to -\infty} \sum_{n=1}^{+\infty} \frac{x^n}{(n+1)^n} = -1$$

So we are missing something.

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Answer 1

I shall derive the asymptotics of this function as $x\to\pm\infty$. Since $$ \frac{1}{{(n + 1)^n }} = \frac{1}{{(n - 1)!}}\int_0^{ + \infty } {t^{n - 1} {\rm e}^{ - (n + 1)t} {\rm d}t} , $$ we immediately obtain $$ \sum\limits_{n = 1}^\infty {\frac{{( \pm x)^n }}{{(n + 1)^n }}} = \pm x\int_0^{ + \infty } {\exp ( - x( \mp t{\rm e}^{ - t} )){\rm e}^{ - 2t} {\rm d}t} $$ for $x>0$. With the upper sign, there is a sole, simple saddle point on the path of integration at $t=1$. Consequently, by the saddle point method $$ \sum\limits_{n = 1}^\infty {\frac{{x^n }}{{(n + 1)^n }}} \sim \sqrt {\frac{{2\pi }}{{{\rm e}^{\rm 3} }}} x^{1/2} \exp (x/{\rm e}) $$ as $x\to+\infty$. With the lower sign, we split the path of integration at $t=1$, and perform the change of variables $t = - W_0 ( - s), - W_1 ( - s)$ using two real branches of the Lambert-$W$ function. Thus, $$ \sum\limits_{n = 1}^\infty {\frac{{( - x)^n }}{{(n + 1)^n }}} = - x\int_0^{1/{\rm e}} {{\mathop{\rm e}\nolimits} ^{ - xs} \left( {\frac{{{\rm e}^{W_0 ( - s)} }}{{1 + W_0 ( - s)}} - \frac{{{\rm e}^{W_1 ( - s)} }}{{1 + W_1 ( - s)}}} \right){\rm d}s} $$ for $x>0$. By $(4.13.5)$ and $(4.13.11)$ $$ \frac{{{\rm e}^{W_0 ( - s)} }}{{1 + W_0 ( - s)}} - \frac{{{\rm e}^{W_1 ( - s)} }}{{1 + W_1 ( - s)}} = 1 + \mathcal{O}(s) $$ as $s\to 0^+$, and by $(4.13.6)$ $$ \frac{{{\rm e}^{W_0 ( - s)} }}{{1 + W_0 ( - s)}} - \frac{{{\rm e}^{W_1 ( - s)} }}{{1 + W_1 ( - s)}} \sim \frac{{\rm 1}}{{{\rm e}\sqrt {2(1 - {\rm e}s)} }} $$ as $s\to 1/\mathrm{e}^-$. Hence, $$ \sum\limits_{n = 1}^\infty {\frac{{( - x)^n }}{{(n + 1)^n }}} = - 1+\mathcal{O}(x^{-1}) $$ as $x\to+\infty$. The existence of $C>0$ follows from this asymptotics.