Prove that $$\sum_{k=0}^{\infty}\left(1+\frac{k}{x}\right)^{-x}$$ is uniformly convergent on $x\in\left[a,\infty\right).$
According to the equality, $$\frac{x}{1+x}<\ln(1+x)$$ we have, \begin{eqnarray*} \left(1+\frac{k}{x}\right)^{-x} &=& \exp\left\{-x\ln\left(1+\frac{k}{x}\right)\right\} \\ &<& \exp\left\{-x\cdot\frac{k}{x+k}\right\} \end{eqnarray*}
How do I control the RHS in order to apply DCT here? Any suggestions?
For $x \in [a,\infty)$ the summand is a monotonically decreasing function of $x$.
Note that for
$$f(x)=\left(1+\frac{k}{x}\right)^{-x}$$
we have
$$\lim_{x \rightarrow \infty}f(x) = 0,\\\lim_{x \rightarrow 0+}f(x) = 1,\\f'(x) = \left(1+\frac{k}{x}\right)^{-x}(k+x)^{-1}\left[k - (k+x)\ln\left(1+\frac{k}{x}\right)\right]\leq 0$$
Hence,
$$\left(1+\frac{k}{x}\right)^{-x}\leq\left(1+\frac{k}{a}\right)^{-a}$$
If $a > 1$, by the integral test
$$\sum_{k=1}^{\infty}\left(1+\frac{k}{a}\right)^{-a} < \infty$$
and
$$\sum_{k=1}^{\infty}\left(1+\frac{k}{x}\right)^{-x}$$
converges uniformly for $1 < a \leq x$.
The series diverges if $x \leq 1$.