Is $$ \sum_{n=0}^\infty {x^n \over \sqrt{n!}} > 0 $$ for all real $x$? (I think it is.) If so, how would one prove this? (To confirm: This is the power series for $e^x$, except with the denominator replaced by $\sqrt{n!}$.)

35$\begingroup$ I don't know why people are voting to close this. I would be interested in learning about methods that could be used to prove that an entire function, given by convergent power series, had no real zeroes, so I would like this question to stay open. $\endgroup$– David E SpeyerJan 5, 2012 at 15:07

5$\begingroup$ I completely agree with @David. $\endgroup$– Igor RivinJan 5, 2012 at 15:33

7$\begingroup$ $1/\sqrt{n!}$ for the absolute value of coefficients is an interesting choice for random polynomials (called Weyl polynomials), their roots are roughly uniformly distributed in a large disc. $\endgroup$– Roland BacherJan 5, 2012 at 17:09

27$\begingroup$ This is very far from a solution, but might be of interest. Let $F(x) = \sum_{n=0}^\infty x^n/\sqrt{n!}$ and let $H(x) = F(x)F(x)$. Then $H(x) = \sum_{n=0}^\infty c_nx^n/\sqrt{n!}$ where $c_n = \sum_{r=0}^n (1)^r \sqrt{\binom{n}{r}}$. It is clear that $c_n = 0$ if $n$ is odd, and it would be sufficient to prove that $c_n > 0$ if $n$ is even, since then $F(x) = H(x)/F(x)$ is a positive function. I have checked this is true for $n \le 100$. $\endgroup$– Mark WildonJan 5, 2012 at 17:34

14$\begingroup$ @Mark Wildon You should really post that observation as a separate question, because it is absolutely bizarre that it works. The largest term in the sum defining $c_{1000}$ is about $1.4 \times 10^{300}$. The fact that you get such nice cancellation to get an answer near $0.02$ seems like a miracle to me. Of course, $\sum (1)^r \binom{n}{r}$ has even larger terms cancelling, but that is for a very good reason; I can't see any reason for your sum to be so small. $\endgroup$– David E SpeyerJan 5, 2012 at 22:45
8 Answers
Looks like the computers really spoiled us :)
GH gave a perfectly valid answer already but the cheapest way to prove positivity is to write $\int_0^1(1t^n)\log(\frac 1t)^{3/2}\,\frac{dt}t=c\sqrt n$ with some positive $c$ (just note that the integral converges and the integrand is positive, and make the change of variable $t^n\to t$). Hence $\int_0^1 (f(x)f(xt))\log(\frac 1t)^{3/2}\,\frac{dt}t=cxf(x)$. If $x$ is the largest zero of $f$ (which must be negative), then plugging it in, we get $0$ on the right and a negative number on the left, which is a clear contradiction. Thus, crossing the $x$axis is impossible. Of course, there is nothing sacred about $1/2$. Any power between $0$ and $1$ works just as well.

3$\begingroup$ Very slick! I had not seen that trick before. $\endgroup$ Jan 6, 2012 at 13:26

4$\begingroup$ Fedja, thank you. Because $d e^x / dx = e^x$, you can show that $e^x$ is positive by asking what would happen at a zero crossing. I've tried (unsuccessfully) to analyze $f'(x)$ with the hope of doing something similar. Your proof strikes me as having the same flavor as that sort of manipulation. Could you offer any motivation for how you came up with your integral "operator"? $\endgroup$ Jan 6, 2012 at 14:35

2

24$\begingroup$ @ J Russell. Yes, I started with the proof for the exponent you mentioned. The main difference is that you need halfderivative here, not the full one. The halfderivative (or, rather, quarterLaplace) operator is wellknown: it is just (regularized) convolution with $t^{3/2}$. We needed the onesided version here (the other side is not integrable), which works especially well on the negative exponents: $\int_0^\infty \frac{e^{nx}e^{n(x+t)}}{t^{1/2}}\frac{dt}{t}=n^{1/2}e^{nx}$, so I just made the logarithmic change of variable to make it act on powers as needed. $\endgroup$– fedjaJan 6, 2012 at 16:53

5$\begingroup$ Nice!! For what its worth, $c$ is $\int_0^\infty (1e^{x}) \phantom. dx^{3/2} = 2 \Gamma(1/2) = 2\sqrt{\pi}$ by integration by parts. [My previous comment, now deleted, reported an incorrect answer because I left the factor $1/t$ out of the integrand in the numerical calculation.] $\endgroup$ Jan 7, 2012 at 23:02
The affirmative answer follows from my response to this related question.
EDIT. Noam Elkies gave a nicer and more general argument here.

$\begingroup$ GH, thanks to you and Noam (and to de Bruijn) for this nice piece of analysis. $\endgroup$ Jan 6, 2012 at 14:34
Here is another nonanswer. In "Asymptotic Methods in Analysis", chapter 6, de Bruijn proves that $$S(s,n)=\frac{2}{\pi}\Gamma(s)(2ns\log 2n)^{s}\left(\sin(\pi s)+O\left((\log n)^{1}\right)\right)$$ where $$S(s,n)= \sum_{k=0}^{2n} (1)^k \binom{2n}{k}^s$$ for all $0\le s\le\frac{3}{2}$. So at least this explains things asymptotically.

$\begingroup$ Actually he proves this for $0<s<\frac{3}{2}$. $\endgroup$ Jan 6, 2012 at 4:22

3$\begingroup$ Inspired by your response I answered the question in the affirmative. See my response. $\endgroup$ Jan 6, 2012 at 5:57
Additional data for Liviu's plots. I used Pari/GP with 1200 digits dec prec, documenting also the required number of terms after which the absolute values of the summands of the series decrease below 1e100. There seems to be no local minimum...
$\small \begin{array}{rlr} & & \text{# of terms}\\ x & f(x) & \text{ required} \\ \hline \\ 1 & 0.438599896749 & 201 \\ 2 & 0.247539616819 & 201 \\ 3 & 0.162554775870 & 211 \\ 4 & 0.117399404501 & 257 \\ 5 & 0.0903120618145 & 304 \\ 6 & 0.0726061182760 & 354 \\ 7 & 0.0602796213492 & 407 \\ 8 & 0.0512783927864 & 464 \\ 9 & 0.0444561508357 & 525 \\ 10 & 0.0391295513879 & 589 \\ 11 & 0.0348689168813 & 658 \\ 12 & 0.0313919770798 & 730 \\ 13 & 0.0285063993737 & 808 \\ 14 & 0.0260770215882 & 889 \\ 15 & 0.0240063146159 & 976 \\ 16 & 0.0222222780410 & 1067 \\ 17 & 0.0206706877888 & 1162 \\ 18 & 0.0193099849974 & 1263 \\ 19 & 0.0181078191003 & 1369 \\ 20 & 0.0170386561852 & 1479 \\ 21 & 0.0160820905671 & 1595 \\ 22 & 0.0152216309789 & 1715 \\ 23 & 0.0144438135509 & 1841 \\ 24 & 0.0137375438980 & 1972 \\ 25 & 0.0130936024884 & 2108 \\ 26 & 0.0125042681404 & 2250 \\ 27 & 0.0119630281606 & 2396 \\ 28 & 0.0114643528377 & 2548 \\ 29 & 0.0110035182996 & 2705 \\ 30 & 0.0105764661081 & 2867 \\ 31 & 0.0101796910429 & 3035 \\ 32 & 0.00981015071575 & 3208 \\ 33 & 0.00946519223932 & 3386 \\ 34 & 0.00914249232841 & 3569 \\ 35 & 0.00884000806032 & 3758 \\ 36 & 0.00855593615550 & 3953 \\ 37 & 0.00828867911422 & 4152 \\ 38 & 0.00803681690505 & 4357 \\ 39 & 0.00779908317617 & 4567 \\ 40 & 0.00757434517200 & 4783 \\ 41 & 0.00736158670179 & 5004 \\ 42 & 0.00715989363457 & 5231 \\ 43 & 0.00696844149585 & 5462 \\ 44 & 0.00678648482039 & 5700 \\ 45 & 0.00661334797911 & 5942 \\ 46 & 0.00644841724806 & 6190 \\ 47 & 0.00629113392871 & 6444 \\ 48 & 0.00614098836080 & 6703 \\ 49 & 0.00599751469633 & 6967 \\ 50 & 0.00586028632445 & 7236 \\ 51 & 0.00572891185489 & 7511 \\ 52 & 0.00560303158255 & 7792 \\ 53 & 0.00548231436720 & 8078 \\ 54 & 0.00536645487311 & 8369 \\ 55 & 0.00525517112099 & 8666 \\ 56 & 0.00514820231209 & 8968 \\ 57 & 0.00504530688991 & 9275 \\ 58 & 0.00494626080983 & 9588 \\ 59 & 0.00485085599129 & 9907 \\ 60 & 0.00475889893049 & 10230 \\ 61 & 0.00467020945455 & 10560 \\ 62 & 0.00458461960073 & 10894 \\ 63 & 0.00450197260623 & 11234 \\ 64 & 0.00442212199624 & 11580 \\ 65 & 0.00434493075923 & 11931 \\ 66 & 0.00427027059992 & 12287 \\ 67 & 0.00419802126157 & 12649 \\ 68 & 0.00412806991028 & 13016 \\ 69 & 0.00406031057475 & 13388 \\ 70 & 0.00399464363573 & 13766 \end{array} $

$\begingroup$ Thanks for the table. After I wrote that there is a minimum I did some additional computations and I realized I was wrong. Your table suggests that the question is far form trivial. $\endgroup$ Jan 6, 2012 at 0:13

2$\begingroup$ Note again the amazing cancellation. To compute that −70 term, the largest term in the sum is $(70)^{70^2}/\sqrt{(70)^2!} \approx e^{70^2/2}$. $\endgroup$ Jan 6, 2012 at 0:23

$\begingroup$ This function is indeed miraculous. $\endgroup$ Jan 6, 2012 at 0:57

1$\begingroup$ Gottfried, thank you for the numerical evaluation. This is consistent with how I believe the series behaves and with the numerical calculations I have done, although I didn't have the tools to push it out anywhere near as far as you did. A comment and an openended question: Presumably one could use interval arithmetic to produce a computeraided proof of positivity for some negative values of $x$. Suppose you could prove positivity down to some large negative value. How large in magnitude would that $x$ have to be for you to really believe, absent a proof, that the series is positive? $\endgroup$ Jan 6, 2012 at 1:50

$\begingroup$ @David Speyer: I'm wondering if the cancellation is more amazing / surprising than the cancellation that happens for the power series for $\exp(x)$ for large negative values of x. $\endgroup$ Sep 13, 2016 at 20:02
This comment serves to record a partial attempt, which didn't get very far but might be useful to others. Following a suggestion of Mark Wildon and Arthur B, define $$f_n(\alpha) := \sum (1)^r \binom{n}{r}^{\alpha}.$$ This is zero for $n$ odd, so we will assume $n$ is even from now on.
Mark Wildon shows that it would be enough to show that $f_n(1/2) \geq 0$ for all $n$. It is easy to see that $f_n(0) = 1$ and $f_n(1)=0$. Arthur B notes that, experimentally, $f_n(\alpha)$ appears to be decreasing on the interval $[0,1]$. If we could prove that $f_n$ was decreasing, that would of course show that $f_n(1/2) > f_n(1) =0$.
I had the idea to break this problem into two parts, each of which appears supported by numerical data:
1. Show that $f_n$ is convex on $[0,1]$.
2. Show that $f'_n(1) < 0$.
If we establish both of these, then clearly $f_n$ is decreasing.
I have made no progress on part 1, but here is most of a proof for part 2. We have $$f'_n(1) = \sum (1)^r \binom{n}{r} \log \binom{n}{r} = \sum (1)^r \binom{n}{r} \left( \log(n!)  \log r! \log (nr)! \right)$$ $$=2 \sum (1)^r \binom{n}{r} \left( \log(1) + \log(2) + \cdots + \log (r) \right)$$ $$=2 \sum (1)^r \binom{n1}{r} \log r.$$ At the first line break, we combined the $r!$ and the $(nr)!$ terms (using that $n$ is even); at the second, we took partial differences once.
This last sum is evaluated asymptotically in this math.SE thread. The leading term is $\log \log n$, so the sum is positive for $n$ large, and $f'_n$ is negative, as desired. The sole gap in this argument is that the math.SE thread doesn't give explicit bounds, so this proof might only be right for large enough $n$.
This answer becomes much more interesting if someone can crack that convexity claim.
Here is a plot of
$$\frac{1}{100}\left(\sum_{k=0}^{16}\frac{x^k}{\sqrt{k!}}\right)$$
on the interval $[4,0]$. (Above I added the terms up to degree $16$.)
Next, is a plot of
$$\frac{1}{100}\left(\sum_{k=0}^{15}\frac{x^k}{\sqrt{k!}}\right)$$
on the interval $[3,0]$. (Above I added the terms up to degree $15$)
This is one strange series.

$\begingroup$ Degree $n=16$ is of little use for $x=10$, for example. The terms increase in size up to about $n=50$ and only then start to decrease. $\endgroup$ Jan 5, 2012 at 20:10

$\begingroup$ @Gerard I know that is why I chose shorter intervals. The function seems to have a local minimum at $2.44$ and the value there is very small $\approx 0.202$. It's a curious function. $\endgroup$ Jan 5, 2012 at 20:30
Another "notyetanswer"...
I've tried another idea. Assume the function f(x) is expressed by the following composition:
$$\small x' = \exp(x)1 $$
$$\small f(x) = g(x') = g(exp(x)1) $$
The idea is, that the unavoidable big "hump" in the partial sums, after which the sequence of partial sums begins to decrease, may be absorbed by the function $\small g(x)$  because $\small \exp(x) $ is really small for large negative x and x' is then very little above 1.
I did not yet arrive at a conclusive result; but the power series for $\small g(x) $ begins with the smooth looking form (and gives the partial sums for $\small x'=\exp(100)1 $):
$\qquad \small
\begin{array} {rr}
\text{powerseries} & \text{partial sums for x' } \\
\hline \\
1.00000000000 & 1.00000000000 \\
+1.00000000000x & 3.72007597602E44 \\
+0.207106781187x^{2} & 0.207106781187 \\
+0.0344748426106x^{3} & 0.172631938576 \\
0.0100670743762x^{4} & 0.162564864200 \\
+0.00821765977664x^{5} & 0.154347204423 \\
0.00654357122833x^{6} & 0.147803633195 \\
+0.00537330847179x^{7} & 0.142430324723 \\
0.00451702185603x^{8} & 0.137913302867 \\
+0.00386915976824x^{9} & 0.134044143099 \\
0.00336528035075x^{10} & 0.130678862748 \\
+0.00296428202807x^{11} & 0.127714580720 \\
0.00263893325448x^{12} & 0.125075647465 \\
+0.00237058888853x^{13} & 0.122705058577 \\
0.00214611388717x^{14} & 0.120558944690 \\
+0.00195602261228x^{15} & 0.118602922077 \\
0.00179331457091x^{16} & 0.116809607506 \\
+0.00165272361723x^{17} & 0.115156883889 \\
0.00153022060566x^{18} & 0.113626663284 \\
+0.00142267593977x^{19} & 0.112203987344 \\
0.00132762563657x^{20} & 0.110876361707 \\
+0.00124310598493x^{21} & 0.109633255722 \\
0.00116753462507x^{22} & 0.108465721097 \\
+0.00109962364925x^{23} & 0.107366097448 \\
\end{array}
$
The the question is, for some large negative x, say $\small x=100 \qquad x'=exp(100)1 = 1+ \epsilon $ the series $\ g(x') $ converges to zero. Unfortunately  although we've translated the original problem to one with nice small numbers I don't see, how to really come nearer a solution, because the convergence of $\small g(1+\epsilon) $ is really slow  if it converges at all to a positive value... So this is not yet a solution, but perhaps a suggestion for a path to try...
Although the following does not provide another proof (perhaps it is possible to attempt one on this basis) I found it nice to see the following pictures.
Let's take from the series $f(x) = \sum_{k=0}^\infty {x^k \over \sqrt{k!}}$ the following variants in the same spirit as we have the hyperbolic and trigonometric series from the exponentialseries:
$$\begin{array}{}
\small \exp_{\tiny \sqrt{\,}}(x) &=& f(x) \\
\small \cosh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k} \over \sqrt{(2k)!}} \\
\small \sinh_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty {x^{2k+1} \over \sqrt{(2k+1)!}} \\
\small \tanh_{\tiny \sqrt{\,}}(x) &=& { \sinh_{\tiny \sqrt{\,}}(x)\over \cosh_{\tiny \sqrt{\,}}(x) } \\
\small \cos_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (1)^k {x^{2k} \over \sqrt{(2k)!}} \\
\small \sin_{\tiny \sqrt{\,}}(x) &=& \sum_{k=0}^\infty (1)^k {x^{2k+1} \over \sqrt{(2k+1)!}} \\
\end{array}$$
The answer to your question is equivalent to say, that always (="for real $x$")
 $\small \cosh_{\tiny \sqrt{\,}}(x)$ is larger than $\small \sinh_{\tiny \sqrt{\,}}(x) $ $\qquad \qquad$ or that
 $\small \mid \tanh_{\tiny \sqrt{\,}}(x) \mid \lt 1$
To illustrate this I've plotted the $\sinh_{\tiny \sqrt{\,}}$ and $\cosh_{\tiny \sqrt{\,}}$curves:
This gives surely an extremely familiar impression...
The $\tanh_{\tiny \sqrt{\,}}$curve looks completely familiar too:
and the image suggests, that indeed the absolute value of $\small \tanh_{\tiny \sqrt{\,}}(x) $ very likely is smaller than $1$ for all real $x$.
However, things are different for the $\sin_{\tiny \sqrt{\,}}$ and $\cos_{\tiny \sqrt{\,}}$ curves  they deviate strongly from the nicely periodic common trigonometric functions:
and combined they do not give a circle, but some ugly thing, strongly distorted (yaxis by $\small \cos_{\tiny \sqrt{ \,} }(\phi)$, xaxis by $\small \sin_{\tiny \sqrt{ \,} }(\phi)$, $\phi$ from $5$ to $+5$) :