bounding a sum using a definite integral Conjecture.  Let $1<p<\infty$.  Then there exists $C\in(0,\infty)$ such that for any $k\in\mathbb{Z}^+$ we have
\begin{equation}\sum_{n=1}^k(k+1-n)^{-\frac{p}{p+1}}n^{-\frac{1}{p+1}}\leq C.\end{equation}
Discussion.  I was thinking of estimating the above sum using a definite integral, like so:
\begin{equation*}\int_1^{k}\frac{dx}{x^{1/(p+1)}(k+1-x)^{p/(p+1)}}\end{equation*}
However, I don't remember any tools from calculus which would enable me to uniformly bound this integral, much less compute it!  Anyone have any ideas?  I suppose I could plug it into MATLAB to see if the conjecture is possibly true.  But even if I do that, I need a rigorous proof.
Thanks guys!
EDIT:  Just to be clear, $C$ is allowed to depend on $p$, but it is not allowed to depend on $k$.
 A: Rephrasing the conjecture: Let $\alpha,\beta>0$ with $\alpha+\beta=1$. Then there exists $C>0$ such that
$$
s_{\alpha,\beta}(k) = \sum_{n=1}^k (k+1-n)^{-\alpha} n^{-\beta} < C
$$
uniformly in $k$.
Split the sum around $n=\frac k2$. For the smaller values of $n$,
$$
\sum_{n=1}^{k/2} (k+1-n)^{-\alpha} n^{-\beta} \le (k/2)^{-\alpha} \sum_{n=1}^{k/2} n^{-\beta} \sim (k/2)^{-\alpha} \frac{(k/2)^{1-\beta}}{1-\beta} < \frac2{1-\beta}.
$$
(The $\sim$ is because $\sum_{n=1}^{k/2} n^{-\beta} \sim \int_{t=1}^{k/2} t^{-\beta}\,dt$.) Similarly, for the larger values of $n$,
$$
\sum_{n=k/2}^k (k+1-n)^{-\alpha} n^{-\beta} \le (k/2)^{-\beta} \sum_{n=k/2}^k (k+1-n)^{-\alpha} = (k/2)^{-\beta} \sum_{m=1}^{k/2} m^{-\alpha} < \frac2{1-\alpha}
$$
by the same computation. Therefore $s_{\alpha,\beta}(k)$ is asymptotically at most $2/(1-\alpha)+2/(1-\beta)$, which implies that there exists $C>2/(1-\alpha)+2/(1-\beta)$ such that $s_{\alpha,\beta}(k) < C$ for every $k$.
[Tangent: if we set $S_{\alpha,\beta}(x) = \sum_{k=1}^\infty s_{\alpha,\beta}(k) x^{k+1}$, then it's easy to check that $S_{\alpha,\beta}(x) = R_\alpha(x)R_\beta(x)$ where $R_\alpha(x) = \sum_{k=1}^\infty k^{-\alpha} x^k$ and similarly for $R_\beta$. (In other words, the sequence $\{s_{\alpha,\beta}(k)\}_{k=1}^\infty$ is naturally the convolution of the sequences $\{k^{-\alpha}\}$ and $\{k^{-\beta}\}$.) Both power series defining $R_\alpha(x)$ and $R_\beta(x)$ have radius of convergence $1$, and both series converge at $x=-1$ by the alternating series test (indeed, they both converge for every complex $z$ with $|z|=1$ except $z=1$).
I would like to then conclude that the power series defining $S_{\alpha,\beta}(x)$ also converges at $x=-1$; this would imply a stronger result, namely that $s_{\alpha,\beta}(k)\to0$ as $k\to\infty$. However, arbitrary rearrangement of power series is only possible when the series converges absolutely, and this is not true for the $R_\alpha(-1)$ and $R_\beta(-1)$ series. So some other justification for the convergence of the $S_{\alpha,\beta}(-1)$ would be necessary.]
