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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.

Susan_Math123
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  • You can't, at least not directly. In a case like that sum you mention it's possible you can show that the $n$-th term is less than or equal to $b_n$, and then use the integral test to show $b_n$ converges. – David C. Ullrich Sep 22 '15 at 17:04
  • Maybe try the ratio test, after converting the binomial coefficient into factorials and simplifying. It seems likely that the ratio from the binomial coefficients would approach 1 and then the division by $2^r$ would make this into $1/2.$ Hmm it seems I was wrong about the ratio of the binomials... – coffeemath Sep 22 '15 at 17:08
  • After some limited experiments it seems the ratio test is inconclusive, since the ratio approaches 1. – coffeemath Sep 22 '15 at 17:16

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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.

coffeemath
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  • Can we find a better upper bound, instead of using $2r \choose r$ as the upper bound for each term? – Susan_Math123 Sep 24 '15 at 03:28
  • @Kazem -- When $r$ gets large compared to $m$ (where $m$ is fixed as in your problem), the binomial $\binom{2r}{r+m}$ becomes relatively close to the middle coefficient $\binom{2r}{r}$ (I don't see immediately a "better" upper bound, but will look more at it.) – coffeemath Sep 24 '15 at 10:11