Let $a_n$ be a real valued sequence. Assume that $a_n\to0$ as $n\to\infty$. Use this to prove that $(1+a_n/n)^n\to1$ as $n\to\infty$.

This proof would be simple if it wasn't for that pesky exponent. I suspect that I"m supposed to use the Bernoulli inequality to solve this somehow, along with monotone convergence theorem. But I'm just really struggling to make any progress with this. Any advice?

  • $\begingroup$ The limit should be $1$, not $0$. It's tending towards $e^0 = 1$. $\endgroup$ – Alex Ortiz Mar 13 '18 at 0:42
  • $\begingroup$ Yes, you're right, sorry, I fixed the typo. $\endgroup$ – jippyjoe4 Mar 13 '18 at 0:46
  • $\begingroup$ This is a marvelous lemma I learnt on this website courtesy Thomas Andrews : see math.stackexchange.com/a/1451245/72031 This is especially useful in the development of the theory of exponential functions based on definition $$\exp(z) =\lim_{n\to\infty} \left(1+\dfrac{z}{n}\right)^{n}$$ $\endgroup$ – Paramanand Singh Mar 13 '18 at 6:47

Let observe that the limit trivially holds for $a_n=0$, thus without loss of generality assume $a_n\neq 0$.

1. Proof by log

Note that

$$(1+a_n/n)^n=e^{n\log(1+a_n/n)}\sim e^{a_n}\to 1$$


$$\log(1+a_n/n)=\frac{a_n}{n}\frac{\log(1+a_n/n)}{a_n/n}\sim \frac{a_n}{n}$$

2. Proof by squeeze theorem

As an alternative note that by Bernoulli and by $\left(1+\frac{1}{n}\right)^{n}\le e$ we have that

$$1+a_n=1+n\frac{a_n}{n}\le \left(1+\frac{a_n}{n}\right)^n\le \left(1+\frac{|a_n|}{n}\right)^n=\left[\left(1+\frac{|a_n|}{n}\right)^{\frac{n}{|a_n|}}\right]^{|a_n|}\le e^{|a_n|}$$

thus by squeeze theorem

$$\left(1+\frac{a_n}{n}\right)^n\to 1$$

  • $\begingroup$ Unfortunately, I cannot use logarithms in this proof. That's why I suspected that I needed to use the Bernoulli inequality and monotone convergence theorem. $\endgroup$ – jippyjoe4 Mar 13 '18 at 0:47
  • $\begingroup$ sorry I didn’t get this request by the OP $\endgroup$ – user Mar 13 '18 at 0:52
  • $\begingroup$ I don’t think that monoticity can help since $a_n$ has not continditions on that, maybe squeeze theorem $\endgroup$ – user Mar 13 '18 at 0:54
  • $\begingroup$ You may have a look at the proof given in linked answer (see comments to question) which uses inequalities and does not need any stuff like $\log$ or $e$. $\endgroup$ – Paramanand Singh Mar 13 '18 at 6:51
  • $\begingroup$ @ParamanandSingh what about the second proof given here by Bernoulli and $(1+1/n)^n<e$ via squeeze theorem? Thanks for the link $\endgroup$ – user Mar 13 '18 at 7:32

(i). For $x\geq 0$ we have $$0\leq \ln (1+x)=\int_1^{1+x}\frac {1}{y}dy\leq \int_1^{1+x} 1dy=x.$$ (ii). For $z\in (-1,0)$ we have $$0> \ln (1-z)=\int_1^{1-z}\frac {1}{y}dy>\int_1^{1-z}\frac {1}{1-z}dz= \frac {-z}{1-z}.$$

(iii). Since $a_n\to 0$ we have $|a_n/n|<1$ for all but finitely many $n.$

(iv). If $a_n\geq 0$ then by (i) we have $$0\leq \ln ((1+a_n/n)^n)=n\ln (1+a_n/n)\leq n(a_n/n)=a_n=|a_n|.$$ (v). If $-1<a_n/n$ then by (ii) we have $$0>\ln ((1+a_n/n)^n)=n\ln (1+a_n/n)=n\ln (1-|a_n/n|)>n\frac {-|a_n/n|}{1-|a_n/n|}=$$ $$=-\frac {|a_n|}{1-|a_n|/n}.$$ (vi). By (iv) and (v), when $|a_n/n|<1$ we have $$|\ln ((1+a_n/n)^n)|\leq \max\left(|a_n|, \frac {|a_n|}{1-|a_n|/n}\right)=\frac {|a_n|}{1-|a_n|/n}.$$ Therefore $\lim_{n\to \infty} \ln ((1+a_n/n)^n)=0.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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