Proving that a sequence is monotone

Solution 1:

I've spent too much time on this already, so I will just dump everything new into another answer.

First,

$\begin{array}\\ c_j &=\binom{-1/2}{j}\\ &=\dfrac1{j!}\prod_{i=0}^{j-1} (-\frac12-i)\\ &=\dfrac{(-1)^j}{j!}\prod_{i=0}^{j-1} (\frac12+i)\\ &=\dfrac{(-1)^j}{2^jj!}\prod_{i=0}^{j-1} (1+2i)\\ &=\dfrac{(-1)^j\prod_{i=0}^{j-1} (2i+1)}{2^jj!}\\ &=\dfrac{(-1)^j\prod_{i=0}^{j-1} (2i+1)\prod_{i=1}^{j} (2i)}{2^jj!\prod_{i=1}^{j} (2i)}\\ &=\dfrac{(-1)^j(2j)!}{2^jj!2^j j!}\\ &=\dfrac{(-1)^j(2j)!}{4^jj!^2}\\ &=\dfrac{(-1)^j}{4^j}\binom{2j}{j}\\ &\approx \dfrac{(-1)^j}{4^j}\dfrac{4^j}{\sqrt{\pi j}}\\ &= \dfrac{(-1)^j}{\sqrt{\pi j}}\\ \end{array} $

Note that $\dfrac{c_{j+1}}{c_j} =\dfrac{\dfrac1{(j+1)!}\prod_{i=0}^{j} (\frac12+i)}{\dfrac1{j!}\prod_{i=0}^{j-1} (\frac12+i)} =\dfrac{j+\frac12}{j+1} =1-\dfrac{1}{2(j+1)} $ so the $c_j$ is decreasing.

Now we can get an infinite converging series for $s_m$.

$\begin{array}\\ s_m &= \sum_{k=1}^{m} \frac{1}{\sqrt{m^2 + k}}\\ &= \frac1{m}\sum_{k=1}^{m} \frac{1}{\sqrt{1 + k/m^2}}\\ &= \frac1{m}\sum_{k=1}^{m} (1 + k/m^2)^{-1/2}\\ &= \frac1{m}\sum_{k=1}^{m} \sum_{j=0}^{\infty} \binom{-1/2}{j}(k/m^2)^j\\ &= \frac1{m}\sum_{k=1}^{m} \sum_{j=0}^{\infty} (-1)^jc_j(k/m^2)^j\\ &= \frac1{m}\sum_{k=1}^{m} \left(1+\sum_{j=1}^{\infty} (-1)^jc_j(k/m^2)^j\right)\\ &= 1+\frac1{m}\sum_{k=1}^{m} \sum_{j=1}^{\infty} (-1)^jc_j(k/m^2)^j\\ &= 1+\frac1{m} \sum_{j=1}^{\infty}\sum_{k=1}^{m} (-1)^jc_j(k/m^2)^j\\ &= 1+\sum_{j=1}^{\infty}(-1)^jc_jm^{-2j-1}\sum_{k=1}^{m} k^j\\ &= 1+\sum_{j=1}^{\infty}(-1)^jc_jp_j(m)/m^{2j+1} \qquad\text{where } p_j(m)=\sum_{k=1}^{m} k^j\\ &= 1+ \sum_{j=1}^{\infty}(-1)^jc_jp_j(m)/m^{2j+1}\\ \end{array} $

Since, for $j=1, 2, 3, ...$ we have $c_j =\frac12, \frac38, \frac{5}{16}, ... $ and $p_j(m) =\frac12 m(m+1), \frac16 m(m+1)(2m+1), \frac14 m^2(m+1)^2 $, we have

$\begin{array}\\ s_m &=1 -\frac{m(m+1)}{4m^3} +\frac{3m(m+1)(2m+1)}{8\cdot 6m^5} -\frac{5m^2(m+1)^2}{16\cdot 4m^7} +...\\ &=1 -\frac{m+1}{4m^2} +\frac{(m+1)(2m+1)}{16m^4} -\frac{5(m+1)^2}{64m^5} +...\\ \end{array} $

Note that since $c_j \approx \frac{1}{\sqrt{\pi j}} $ and $p_j(m) \approx \frac{m^{j+1}}{j+1} $ we have $c_jm^{-2j-1}p_j(m) \approx \dfrac{1}{m^{j}(j+1)\sqrt{\pi j}} $, so the series converges absolutely.

From this, we can get an exact equation for $s_{m+1}-s_m $.

We first will use the expression with the first 4 terms.

$\begin{array}\\ s_{m+1}-s_m &=(1 -\frac{m+2}{4(m+1)^2} +\frac{(m+2)(2m+3)}{16(m+1)^4} -\frac{5(m+2)^2}{64(m+1)^5} +...)\\ &\qquad -(1 -\frac{m+1}{4m^2} +\frac{(m+1)(2m+1)}{16m^4} -\frac{5(m+1)^2}{64m^5} +...)\\ &=-(\frac{m+2}{4(m+1)^2}-\frac{m+1}{4m^2}) +(\frac{(m+2)(2m+3)}{16(m+1)^4}-\frac{(m+1)(2m+1)}{16m^4})\\ &\qquad -(\frac{5(m+2)^2}{64(m+1)^5}-\frac{5(m+1)^2}{64m^5}) +...\\ &=\frac{m^2 + 3 m + 1}{4 m^2 (m + 1)^2} -\frac{4 m^5 + 19 m^4 + 30 m^3 + 20 m^2 + 7 m + 1}{16 m^4 (m + 1)^4}\\ &\qquad +\frac{5 (3 m^6 + 17 m^5 + 35 m^4 + 35 m^3 + 21 m^2 + 7 m + 1)}{64 m^5 (m + 1)^5}\\ &=\frac{16m^3(m+1)^3(m^2 + 3 m + 1) -4m(m+1)(4 m^5 + 19 m^4 + 30 m^3 + 20 m^2 + 7 m + 1)+5 (3 m^6 + 17 m^5 + 35 m^4 + 35 m^3 + 21 m^2 + 7 m + 1)}{64 m^5 (m + 1)^5}\\ &=\frac{4 m (m + 1) (4 m^6 + 16 m^5 + 13 m^4 - 10 m^3 - 16 m^2 - 7 m - 1)+5 (3 m^6 + 17 m^5 + 35 m^4 + 35 m^3 + 21 m^2 + 7 m + 1)}{64 m^5 (m + 1)^5}\\ &=\frac{m^5 (16 m^3 + 80 m^2 + 131 m + 52)+O(m^4)}{64 m^5 (m + 1)^5}\\ &=\frac{16 m^3 + 80 m^2 + 131 m + 52)+O(1/m)}{64 (m + 1)^5}\\ &\gt\frac{16 (m+1)^3+O(1/m)}{64 (m + 1)^5}\\ &\gt\frac{1+O(1/m^4)}{4(m + 1)^2}\\ \end{array} $

Next, look at all the terms.

$\begin{array}\\ s_{m+1}-s_m &= \sum_{j=1}^{\infty}(-1)^jc_j\left(\dfrac{p_j(m+1)}{(m+1)^{2j+1}}-\dfrac{p_j(m)}{m^{2j+1}}\right)\\ &= \sum_{j=1}^{\infty}(-1)^jc_j\left(\dfrac{p_j(m+1)m^{2j+1}-p_j(m)(m+1)^{2j+1}}{(m+1)^{2j+1}m^{2j+1}}\right)\\ &= \sum_{j=1}^{\infty}(-1)^jc_j\left(\dfrac{d_j(m)}{(m+1)^{2j+1}m^{2j+1}}\right)\\ \end{array} $

where $d_j(m)=p_j(m+1)m^{2j+1}-p_j(m)(m+1)^{2j+1} $.

$\begin{array}\\ d_j(m) &=p_j(m+1)m^{2j+1}-p_j(m)(m+1)^{2j+1} \\ &=(p_j(m)+(m+1)^j)m^{2j+1}-p_j(m)(m+1)^{2j+1}\\ &=(p_j(m)+(m+1)^j)m^{2j+1}-p_j(m)(m+1)^{2j+1} \\ &=p_j(m)(m^{2j+1}-(m+1)^{2j+1})+(m+1)^jm^{2j+1}\\ \\ &=p_j(m)m^{2j+1}(1-(1+1/m)^{2j+1})+(m+1)^jm^{2j+1}\\ \\ &=m^{2j+1}(p_j(m)(1-(1+1/m)^{2j+1}))+(m+1)^j)\\ &\lt m^{2j+1}(p_j(m)(1-(1+(2j+1)/m)))+(m+1)^j)\\ &= m^{2j+1}(p_j(m)(-\frac{2j+1}{m})+(m+1)^j)\\ &\approx m^{2j+1}(-m^{j+1}\frac{2j+1}{m(j+1)}+(m+1)^j) \qquad\text{since }p_j(m)>n^{j+1}/(j+1)\\ &= m^{2j+1}(-m^{j}\frac{2j+1}{(j+1)}+(m+1)^j)\\ &= m^{3j+1}(-\frac{2j+1}{j+1}+(1+1/m)^j)\\ &\approx m^{3j+1}(-2+\frac1{j+1}+(1+1/m)^j)\\ \end{array} $

so if $(1+1/m)^j \lt 2-\frac1{j+1} $ then $d_j(m) < 0$.

This is the same as $1+1/m \lt (2-\frac1{j+1})^{1/j} $ or $m \gt \frac1{(2-\frac1{j+1})^{1/j}-1} $.

Numerically, according to Wolfy, this looks like about $3j/2$. To check:

$\begin{array}\\ (2-\frac1{j+1})^{1/j} &=2^{1/j}(1-\frac1{2j+2})^{1/j}\\ &=2^{1/j}e^{\ln(1-1/(2j+2))/j}\\ &\approx 2^{1/j}e^{-1/(j(2j+2))}\\ &=e^{\ln 2/j-1/(j(2j+2))}\\ &=e^{\ln 2/j-1/(j(2j+2))}\\ &\approx 1+\ln 2/j-1/(j(2j+2))\\ \text{so}\\ \frac1{(2-\frac1{j+1})^{1/j}-1} &\approx \frac1{\ln 2/j-1/(j(2j+2))}\\ &= \frac{j}{\ln 2-1/(2j+2)}\\ \text{and}\\ \frac1{\ln 2} &\approx 1.44\\ \end{array} $

Experiments with Wolfy suggest that a more accurate approxumation is $\frac{j}{\ln 2-1/(3.85j)} $.

Estimating the terms.

Each term in the sum is about

$\begin{array}\\ \dfrac{d_j(m)}{(m+1)^{2j+1}m^{2j+1}} &\approx \dfrac{m^{3j+1}(-1+\frac1{j+1}+\frac{j}{m})}{(m+1)^{2j+1}m^{2j+1}}\\ &\approx -\dfrac{m^j(1-\frac1{j+1})}{(m+1)^{2j+1}}\\ &\approx -\dfrac{1-\frac1{j+1}}{m^{j+1}}\\ \end{array} $

Taking $c_j$ into account, this is about $-\dfrac{c_j(1-\frac1{j+1})}{m^{j+1}} $

This is certainly decreasing in absolute value, so the sum will be between the last two sums.