Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Usually when working with indefinite sums, I want to work out the sum or whether it convergence. But now I encountered a problem they other way around and I'm clueless...

Is there even a general solution of $$ x=\sum_{n=0}^\infty e^{-A_n/x}$$ for $A_n$, where $x$ is given and real, $A_n >0\space\forall n$ and $\frac{dA_n}{dx}=0\space\forall n$?

Thank you


To make my question clearer for the commenters and others, I'm searching for a systematic sequence $A_n$ which, when entered in the equation above, yields $x$ and this should hold for all (real) $x$.

share|cite|improve this question
You want to solve for $A_n$? That seems to be one equation in an infinite number of unknowns. To have convergence, you need the $A_n$ to increase fast enough, but once you do, you can bump $A_1$ up and $A_2$ down and maintain the equality. I'm not sure to make of the condition $\frac{dA_n}{dx}=0\space\forall n$ – Ross Millikan Jul 10 '11 at 1:01
Does not make much sense to me. You could, for example set $a_n = b (n+1)$ – leonbloy Jul 10 '11 at 1:15
@Ross, the condition $\frac{dA_n}{dx}=0$ only means $A_n$ does not depend on $x$. I state this for completeness, otherwise a trivial solution could be found, e.g $A_n(x)=-x(log(x)+e^{π2/6n^2})$. @leonbloy: That $A_n$ is not a solution of the equation. – JBSnorro Jul 10 '11 at 2:08
But what does $A_n$ does not depend on $x$ mean? Given a certain solution sequence $A_n$, either it satisfies the equation for a fixed $x$ (and then it is implicitly a function of $x$ : different $x$ give different $A_n$), either it satisfies the equation for all $x$ (I'd like to see that) – leonbloy Jul 10 '11 at 3:56
Clearly the right-hand side is always positive, so this can only happen for $x>0$. @Ross, you answer shows that this can always be arranged so that the series $\sum_{n=0}^\infty e^{-A_m/x}$ converges to a function $x$ for at least one point; I think the interesting question that's hiding in here is whether it's possible for the series to converge to $x$ for $0<a<x<b$ for some interval $(a,b)$ in the positive reals. – Vladimir Sotirov Jul 10 '11 at 6:47
up vote 6 down vote accepted

It cannot be done. For the proof write $x:={1\over y}$. Then we should have $${1\over y}\ \equiv\ \sum_{n=0}^\infty e^{-A_n y}\qquad(*)\ ,$$ say for all $y\geq1$. In particular $\sum_{n=0}^\infty e^{-A_n}=1$, so necessarily $\lim_{n\to\infty} A_n=\infty$. It follows that $\alpha:=\inf_n A_n>0$ and therefore $$\sum_{n=0}^\infty e^{-A_n y}=\sum_{n=0}^\infty e^{-A_n} \ e^{-A_n(y-1)} \leq e^{-\alpha(y-1)} \qquad (y\geq1)\ .$$ This shows that $(*)$ cannot hold for all $y\geq 1$.

share|cite|improve this answer
Brilliant. This in fact shows that if $\frac 1y=\sum_{n=0}^\infty e^{-A_ny}$ and $\frac1x=\sum_{n=0}^\infty e^{-A_nx}$ with some $y>x$, then $\frac xy\leq e^{-\alpha(y-x)}$, limiting the intervals on which we could have convergence (because when $x/y\to0$ slower than $e^{-\alpha(y-x)}$ as $y\to\infty)$. Some numerical exploration with Mathematica suggests that for $\alpha\geq 1$, $x\geq 1$ the above system has no solution other than $x=y$, which seems to follow by computing derivatives $-x/y^2>-1>-\alpha e^{-\alpha(y-x)}$. – Vladimir Sotirov Jul 10 '11 at 18:57

Certainly there are many solutions. As $x$ grows, the right side decreases monotonically, so for any series $A_n$ that is convergent there will be an $x$ that solves the equation. So pick any $x$ and series $A_n$ that solve the problem. Given a different $x$, just change your favorite $A_n$(s) to make it work.

For a specific example, take $x=1, A_n=\ln 2^{n+1}$. If you want a solution for $x=2$, just decrease any set of $A_n$'s to add enough to the RHS.

share|cite|improve this answer
Two remarks on your post. First, a necessary condition for the RHS to be finite is that $A_n\to+\infty$. Second, more importantly, the RHS is nondecreasing. – Did Jul 10 '11 at 3:12
@Didier: I remarked on the need for $A_n$ to increase in my comment on the original question. I was more focused on the fact that there are so many solutions the problem is not too interesting. – Ross Millikan Jul 10 '11 at 3:17
But, of course, this implies that $A_n$ is implicitly a function of $x$ – leonbloy Jul 10 '11 at 3:53
@Ross: My remarks addressed your post and your post only, and specifically your As $x$ grows, the right side decreases monotonically and for any series $A_n$ that is convergent. I can understand neither of them. First, as $x$ grows, the RHS increases. Second, for any sequence $(A_n)$ such that $A_n\not\to+\infty$ and any fixed positive $x$, $\mathrm{e}^{-A_n/x}\not\to0$ when $n\to+\infty$, hence the RHS is infinite. – Did Jul 10 '11 at 9:26
@Ross: Not interested in clarifying the above? – Did Jul 21 '11 at 21:41

Divide both sides by $x$ to get:
$$1 = \Sigma\frac{e^{-A_n/x}}{x},$$ As $x\to \infty$, the left hand side is 1 while the right hand side goes to zero for each $n$.

This might give a hint on what $A_n$ won't work.

share|cite|improve this answer
It seems that what's going on here is that we're trying to write $1/x$ as a Fourier transform but where the transform is restricted to involve only a countable number of frequencies $A_n$. That's clearly impossible. – Carl Brannen Jul 10 '11 at 5:42
you're right that what's happening is writing $-1/x$ as a series in $\{e^{mx}\}$ for positive real $m$. I agree that this is impossible when we require the series to converge absolutely, but I'm not sure that we have impossibility under conditional convergence (to a non-holomorphic function). – Vladimir Sotirov Jul 10 '11 at 6:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.