How can I use the Integral test for convergence when the function under the summation is not monotonically decreasing? For example, I am looking for an upper bound for the following sum in which the function is uni-modal:
$ F= \sum_{r=k+1}^{\infty} \frac{m}{r 2^r} {r \choose \frac{m+r}{2}}$
where $k \geq m$ . Ignore the terms under the sum in which $m$ and $r$ do not have the same parity.
Suppose $m$ is even, so that your sum reqires also $r$ even because of the parity condition. Then we may replace $m$ by $2m$ and $r$ by $2r,$ and then the $r$th term becomes $$\frac{m}{r \cdot 4^r} \cdot \binom{2r}{r+m}. \tag{1}$$ Here the binomial coefficient is bounded above by the middle coefficient $\binom{2r}{r},$ and we may ignore the constant factor $m$ since it isn't summed over.
Now we apply the fact that as $r \to \infty$ the expression $$[\binom{2r}{r}/(4^r)] \cdot \sqrt{r}$$ approaches $1/\sqrt{\pi}.$ With the extra division by $r$ occurring in (1) this means we have an upper bound on the $r$th term of the form $K\cdot r^{-3/2},$ and now may either apply the integral test or just quote the "$p$ - test" to conclude convergence.