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I am looking for an upper bound of the following sum $$ S_p:= \sum_{j=1}^p \frac{p+1}{p-j+1} \frac1{2^j}. $$ The upper bound should be independent of $p$, of course. Numerical experiments indicate that $$ S_p \le \frac53 $$ with the maximum attained for $p=3,4$. However, I am not able to prove it.

I could only prove $S_p \le 3$. Here is, what I did: First, we see $$ \frac{p+1}{p-j+1} = 1 + \frac j{p-j+1}\le 1+j, $$ hence $$ S_p \le \sum_{j=1}^p (1+j) \frac1{2^j} . $$ Using the power series $$ (1-x)^{-2} = \sum_{j=0}^\infty (1+j) x^j , \ |x|<1, $$ I can estimate $$ S_p \le (1-\frac12)^{-2} -1 = 3. $$ How can the bound be improved?

daw
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2 Answers2

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First, note that that the change of summation index $j\leftarrow p+1-j$ shows that $$ S_p=\frac{p+1}{2^{p+1}}\sum_{j=1}^p\frac{2^j}{j} $$ Thus $$ \frac{2^{p+1}}{p+1}S_p=\sum_{j=1}^p\frac{2^j}{j}=\frac{2^p}{p}+\frac{2^{p}}{p}S_{p-1} $$ Similarly$$\eqalign{ \frac{2^{p+2}}{p+2}S_{p+1} &=\frac{2^{p+1}}{p+1}+\frac{2^{p+1}}{p+1}S_{p}\cr &=\frac{2p}{p+1}\left(\frac{2^{p+1}}{p+1}S_p-\frac{2^{p}}{p}S_{p-1}\right)+\frac{2^{p+1}}{p+1}S_{p} }$$ This can be rearranged as follows $$\eqalign{ \frac{S_{p+1}}{S_p} &= \frac{(p+2)(3p+1)}{2(p+1)^2}-\frac{p+2}{2(p+1)}\cdot\frac{S_{p-1}}{S_p} }$$ Now, If for some $p\ge2$ we have $S_p<S_{p-1}$ then $$ \frac{S_{p+1}}{S_p}<\frac{(p+2)(3p+1)}{2(p+1)^2}-\frac{p+2}{2(p+1)}=\frac{p(p+2)}{(p+1)^2}<1. $$ That is $S_{p+1}<S_p$, and we have proved that $$ S_p<S_{p-1}\quad \Longrightarrow\quad S_{p+1}<S_p $$ Now it is easy to check that $S_5=\frac{8}{5}<\frac{5}{3}=S_4$, hence the sequence $(S_p)_p$ is decreasing starting from $p=4$. Since $$S_1=1~<S_2=\frac{3}{2}<S_3=\frac{5}{3}=S_4=\frac{5}{3}$$ we conclude that $$\max\{S_p:p\geq 1\}=\frac{5}{3}$$ and it is exactly attained at $p=3,4$.$\qquad\square$

Remark. as Did suggested, we have also the simple lower bound $$ S_p\geq\frac{p+1}{2^{p+1}}\cdot\frac{2^{p+1}-2}{p} $$ So, $\ell~{\buildrel{\rm def}\over =}~\lim\limits_{p\to\infty}S_p\geq 1$. This implies that $$\inf\{S_p:p\geq 1\}=1=S_1.$$ Finally, with a little more work, it can be proved that $\ell=1$.

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Omran Kouba
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  • Nice. Although the OP does not ask for lower bounds, you might want to add that $S_p\to1$ when $p\to\infty$ hence $\min{S_p;p\geqslant1}=1$ and $\inf{S_p;p\geqslant p_0}=1$ for every $p_0\geqslant1$. – Did Jul 15 '14 at 10:09
  • @Did, Yes you are right, It is done. – Omran Kouba Jul 15 '14 at 10:54
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Hint: Rewrite the rational function $\frac{p+1}{p+1-j}$ as an infinite geometric series, i.e.,

$$\frac{p+1}{p+1-j}=\frac{1}{1-\left(\frac{j}{p+1}\right)}=\sum_{k=0}^{\infty}\left(\frac{j}{p+1}\right)^k.$$

This expansion is valid since the variable $j$ only ranges from $0\le j \le p<p+1$, hence $|\frac{j}{p+1}|<1.$

Then, by reversing the order of summation we get:

$$S_p=\sum_{j=1}^{p}\sum_{k=0}^{\infty}\left(\frac{j}{p+1}\right)^k\frac{1}{2^j}\\ =\sum_{k=0}^{\infty}\sum_{j=1}^{p}\left(\frac{j}{p+1}\right)^k\frac{1}{2^j}\\ =\sum_{k=0}^{\infty}\frac{1}{(p+1)^k}\sum_{j=1}^{p}\frac{j^k}{2^j}.$$

David H
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  • I do not see how to conclude: Using the estimate $j^k\le p^k$, I obtain $S_p \le \frac12 (p+1)$ ?? – daw Jul 15 '14 at 07:40