Does $\lim_{n\to \infty} \sum_{i=1}^n (A^{i/(n+1)}-A^{i/n}) = 0$ for any $A\ge 1$? I conjecture that the following limit
$$\lim_{n\to \infty} \sum_{i=1}^n (A^{i/(n+1)}-A^{i/n})$$
is zero for any $A\ge 1$, but I cannot come up with a proof nor a counterexample, and I was wondering if someone else could.

For some context, I became interested in the limit after reading Greg Martin's comment in this post.
 A: Let $A > 0$. By the mean value theorem, for each $i = 1, \ldots, n$, we have
$$ A^{\frac{i}{n+1}} - A^{\frac{i}{n}} = -\frac{i}{n(n+1)}A^{x_i} \log A $$
for some $x_i \in (\frac{i}{n+1}, \frac{i}{n}) \subseteq [\frac{i}{n+1}, \frac{i+1}{n+1}]$. So,
$$ \sum_{i=1}^{n} (A^{\frac{i}{n+1}} - A^{\frac{i}{n}})
= -\frac{n+1}{n} \sum_{i=1}^{n} \frac{i}{n+1} A^{x_i}(\log A) \frac{1}{n+1}  $$
Regarding the last sum as a Riemann sum, it therefore follows that
$$ \lim_{n\to\infty} \sum_{i=1}^{n} (A^{\frac{i}{n+1}} - A^{\frac{i}{n}})
= -\int_{0}^{1} x A^x \log A \, \mathrm{d}x
= \frac{A - 1}{\log A} - A. $$
(Here, we regard $\frac{A - 1}{\log A} = 1$ if $A = 1$.)
A: No; in fact, it is always negative for $A>1$. We explicitly evaluate the limit. One has
$$A^{\frac{i}{n+1}}-A^{\frac in}=A^{\frac in}\left(A^{-\frac{i}{n(n+1)}}-1\right)=A^{\frac in}\left(e^{\frac{-i\log A}{n(n+1)}}-1\right).$$
Note that $1+x\leq e^x\leq 1+x+x^2$ for $x\leq 0$, so as long as $n\geq \log A$,
$$\left|\left(A^{\frac{i}{n+1}}-A^{\frac in}\right)-A^{\frac in}\left(-\frac{i\log A}{n(n+1)}\right)\right|\leq A^{\frac in}\frac{i^2\log^2A}{n^2(n+1)^2}.$$
This means that
$$\sum_{i=1}^n \left(A^{\frac{i}{n+1}}-A^{\frac in}\right)\text{ and }-\sum_{i=1}^nA^{\frac in}\frac{i\log A}{n(n+1)}$$
differ by at most
$$\sum_{i=1}^n A^{\frac in}\frac{i^2\log^2A}{n^2(n+1)^2}\leq \frac{A\log^2 A}{n^2(n+1)^2}\sum_{i=1}^n n^2=O\left(\frac 1n\right)$$
for large $n$, and so
$$\lim_{n\to\infty}\sum_{i=1}^n \left(A^{\frac{i}{n+1}}-A^{\frac in}\right)=-\log A\lim_{n\to\infty}\frac1{n+1}\sum_{i=1}^n\frac inA^{i/n}.$$
The limit can be interpreted as a Riemann sum for the function $xA^x$, and so the desired expression is
$$-\log A\int_0^1 xA^xdx=-\frac{1+A\log A-A}{\log A}=-A+\frac{A-1}{\log A}.$$
A: We can obtain much more than the limit itself
$$S_n=\sum_{i=1}^nA^{\frac{i}{n+1}} - A^{\frac{i}{n}} =-\frac{A-1}{A^{\frac{1}{n}}-1}+\frac{A-1}{A^{\frac{1}{n+1}}-1}-A$$
Assuming $A>1$ and using series expansions
$$S_n=\frac{A-1}{\log (A)}-A-\frac{(A-1) \log (A)}{12 n^2}\Bigg[1-\frac 1n +\frac{20-\log ^2(A)}{20 n^2}+\frac{\log ^2(A)-10}{10 n^3}+O\left(\frac{1}{n^4}\right)\Bigg]$$
Trying for $A=e$, the truncated series leads to
$$S_{10}=-\frac{5995457+4543 e}{6000000}=  \color{red}{-1.001301}03$$ while, converted to decimals, the exact value is $\color{red}{-1.00130114}$
