Bounding a complicated sequence I ecountered the following horrible sequence during an inductive proof of a bound for another sequence:
$$
\sum\limits_{k = 0}^n {\frac{{(n + 3)^{1.47} }}{{(k + 1)^{1.47} (n - k + 1)^{1.47} }}} 
$$
for $n\geq 0$. I would like to show that this sequence is bounded by $5.63$. It seems to have a maximum at $n=3$ and that it is increasing for $n\geq 89$. Using the symmetry about $n/2$ and Tannery's theorem, I was able to show that it converges to $2\zeta (1.47) = 5.475994 \ldots$. Thus, the bound is true for $n$ large enough. It would be sufficient to show that the bound holds after some explicit value of $n$ and finish the proof by computing the first several terms numerically. Any comments and/or suggestions are welcome.
 A: Here's an approach to a slightly simplified version of the problem; hopefully it still conveys the general "divide and conquer" approach—if trivial bounds don't suffice, divide the summation up into regions and try trivial bounds on each region separately.
We start by writing
$$
2 \sum_{k=1}^{n/2} \frac{n^{1.47}}{k^{1.47}(n-k)^{1.47}} = 2 \sum_{k=1}^{\delta n} \frac1{k^{1.47}} \bigg(\frac{n}{n-k}\bigg)^{1.47} + 2 \sum_{k=\delta n+1}^{n/2} \frac1{k^{1.47}} \bigg(\frac{n}{n-k}\bigg)^{1.47}
$$
for some real number $\delta\in(0,\frac12)$. In the first sum, the second factor is close to constant, so maybe a trivial bound suffices:
\begin{align*}
2 \sum_{k=1}^{\delta n} \frac1{k^{1.47}} \bigg(\frac{n}{n-k}\bigg)^{1.47} &\le 2 \sum_{k=1}^{\delta n} \frac1{k^{1.47}} \bigg(\frac{n}{n-\delta n}\bigg)^{1.47} \\
&= \frac2{(1-\delta)^{1.47}} \sum_{k=1}^{\delta n} \frac1{k^{1.47}} < \frac{2\zeta(1.47)}{(1-\delta)^{1.47}}.
\end{align*}
In the second sum, each term remains the same order of magnitude as $k$ varies, and so trivial bounds yield
\begin{align*}
2 \sum_{k=\delta n+1}^{n/2} \frac1{k^{1.47}} \bigg(\frac{n}{n-k}\bigg)^{1.47} &\le 2 \sum_{k=\delta n+1}^{n/2} \frac1{(\delta n)^{1.47}} \bigg(\frac{n}{n-n/2}\bigg)^{1.47} \\
&= \frac{2^{2.47}}{(\delta n)^{1.47}} \big(\tfrac12n-\delta n \big) < \bigg( \frac2{\delta n} \bigg)^{1.47}.
\end{align*}
We have shown that
$$
2 \sum_{k=1}^{n/2} \frac{n^{1.47}}{k^{1.47}(n-k)^{1.47}} < \frac{2\zeta(1.47)}{(1-\delta)^{1.47}} + \bigg( \frac2{\delta n} \bigg)^{1.47},
$$
which will be small enough if $\delta$ is chosen appropriately small (for the first term on the right-hand side)—in this case $\delta=0.01$ seems good enough—and if $n$ is restricted to be sufficiently large in terms of $\delta$ (for the second term)—in this case $n>1200$ seems good enough.
