Consider for $x>0$ and $k$ a positive integer :
$$ f(x,k) = \sum_{n=1}^{k} (\frac{x}{n})^n = \sum_{n=1}^{k} \frac{x^n}{n^n} $$
and
$$ f(x) = \sum_{n=1}^{\infty} (\frac{x}{n})^n = \sum_{n=1}^{\infty} \frac{x^n}{n^n} $$
Now clearly $f(x) < \exp(x) -1$.
But the approximation $\exp(x/e) -1$ is not too bad.
This follows for instance from stirlings formula.
In fact considering $f(x,k)$ is an analogue to considering the truncated $\exp(x)$ :
But there is something interesting here :
For sufficiently large integer $x$ we have
$$ \exp(-1) - \frac{1}{x - \ln(1+x)} < f(x) - f(x,x) < \exp(-1) - \frac{1}{1+x}$$
or equivalently because
$$f(x) - f(x,x) = \sum_{n=x+1}^{\infty} (\frac{x}{n})^n = \sum_{n=1}^{\infty} (\frac{x}{x+1+n})^{x+1+n} $$ ,
We have conjecture $A$ :
$$ \exp(-1) - \frac{1}{x - \ln(1+x)} < \sum_{n=1}^{\infty} (\frac{x}{x+1+n})^{x+1+n} < \exp(-1) - \frac{1}{1+x}$$
How to prove this ?
How many ways are there to prove this ?
Is using Stirling's approximation neccessary ?
