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

Suppose $${a_n^{s_n} + b_n^{s_n} =1}$$ $$\lim_{n \to \infty} a_n = a >0$$ $$\lim_{n \to \infty} b_n = b >0$$ and $${a^s + b^s =1}$$ where $a+b<1$. Then how can we prove that $\displaystyle\lim_{n \to \infty} s_n = s$?

share|cite|improve this question
There is a $\lim$ on the first relation? – enzotib Aug 14 '12 at 21:07
This is not true if eg $a=a_n=1$ and $b=b_n=0$ so you are missing some conditions. Once we get those special cases out of the way, have you tried thinking what happens if the limit is greater than, or less than $s$ - treat the two cases separately. – Mark Bennet Aug 14 '12 at 21:10
Satisfied with an answer below? – Did Sep 19 '12 at 17:31
up vote 3 down vote accepted

Here’s a more routine approach.

If $\lim_{n\to\infty}s_n\ne s$, then either $\limsup_n s_n>s$, or $\liminf_n s_n<s$.

If $\limsup_n s_n>s$, we may assume (by passing to a subsequence if necessary) that there is some $u>s$ such that $s_n\ge u$ and $1>a_n,b_n>0$ for all $n$. Then for all $n$ we have $a_n^{s_n}+b_n^{s_n}\le a_n^u+b_n^u$, so


which is absurd.

Similarly, if $\liminf_n s_n<s$, we may assume that there is some $u<s$ such that $s_n\le u$ and $a_n,b_n>0$ for all $n$, and we then have


which again is absurd. Thus, $\lim_{n\to\infty}s_n=s$.

share|cite|improve this answer
Please reverse some inequalities: if $u\gt s$, then $a^u\lt a^s$ for $a$ in $(0,1)$. – Did Aug 16 '12 at 10:35
@did: Argh. This not my day. Thanks. – Brian M. Scott Aug 16 '12 at 10:40
@did: WLOG they are in $(0,1)$, since $a,b>0$ and $a+b<1$. I chose not to go through the details of passing to a subsequence, but I did explicitly say that I was doing so. – Brian M. Scott Aug 16 '12 at 10:46

An elementary proof is based on the following not-so-usual way to express the convergence to any positive limit not equal to $1$.

Consider for example the fact that $\lim\limits_{n\to\infty}a_n=a$. Since $a\gt0$, $b\gt0$ and $a+b\lt1$, one knows that $0\lt a\lt1$, hence the function $\alpha:x\mapsto\log(x)/\log(a)$ is continuous and decreasing on $x\gt0$. In particular, the fact that $\lim\limits_{n\to\infty}\alpha(a_n)=\alpha(a)=1$ implies that, for every $0\lt\varepsilon\lt1$, there exists some finite $n_\varepsilon$ such that, for every $n\geqslant n_\varepsilon$, $1-\varepsilon\leqslant\alpha(a_n)\leqslant1+\varepsilon$, that is, $a^{1+\varepsilon}\leqslant a_n\leqslant a^{1-\varepsilon}$.

Likewise, one can, and we will, assume that $b^{1+\varepsilon}\leqslant b_n\leqslant b^{1-\varepsilon}$ for every $n\geqslant n_\varepsilon$. Since $a_n^{s_n}+b_n^{s_n}=1$ for every $n$, one gets, for every $n\geqslant n_\varepsilon$, $$ a^{(1+\varepsilon)s_n}+b^{(1+\varepsilon)s_n}\leqslant1\leqslant a^{(1-\varepsilon)s_n}+b^{(1-\varepsilon)s_n}. $$ Since $a^s+b^s=1$ and the function $r\mapsto a^r+b^r$ is decreasing, $(1-\varepsilon)s_n\leqslant s\leqslant(1+\varepsilon)s_n$ for every $n\geqslant n_\varepsilon$, that is, $\frac{s}{1+\varepsilon}\leqslant s_n\leqslant\frac{s}{1-\varepsilon}$. Finally, $\lim\limits_{n\to\infty}s_n=s$.

share|cite|improve this answer

Consider the function $$f(x,y,t):=x^t+y^t-1\quad(0<x<1,\ 0<y<1,\ t\in{\mathbb R})\ .$$ By assumption $(a,b,s)\in{\rm dom}(f)$ and $f(a,b,s)=0$. Furthermore ${\partial f\over\partial t}=\log x\ x^t+\log y\ y^t$, whence $$\left.{\partial f\over\partial t}\right|_{(a,b,s)}=\log a\ a^s+\log b\ b^s<0\ .$$ It follows by the implicit function theorem that there is an open box $W=U\times V\subset{\mathbb R}^3$ with center $(a,b,s)$ and a $C^1$-function $$\psi:\quad U\to V,\quad (x,y)\to t=\psi(x,y)\ ,$$ such that within this box the equation $f(x,y,t)=0$ is equivalent to $t=\psi(x,y)$.

Coming back to the original question we now have $$\lim_{n\to\infty} s_n=\lim_{n\to\infty} \psi(a_n,b_n)=\psi(a,b)=s\ .$$

share|cite|improve this answer

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.