Cauchy's First Limit Theorem

If limnan=0\displaystyle \lim_{n \to \infty} a_n = 0, then limna1+a2+a3+ann=0\displaystyle \lim_{n \to \infty} \frac{a_1 + a_2 + a_3 + \dots a_n}{n} = 0

As every convergent sequence in R\mathbb{R} is bounded, so there exists positive real number KK such that an<K for all nN|a_n| < K \text{ for all } n \in \mathbb{N}

As limnan=0\displaystyle \lim_{n \to \infty} a_n = 0, so if ϵ>0\epsilon > 0 be given, corresponding to ϵ\epsilon, there exists natural number pp such that an0<ϵ2 for all np|a_n - 0| < \dfrac{\epsilon}2 \text{ for all } n \geq p

a1+a2+a3+ann0a1+a2+a3+apn+ap+1+ap+2+ann<pkn+ϵ2n(np)<pkn+ϵ2 for all np.\Bigg| \dfrac{a_1 + a_2 + a_3 + \dots a_n}{n} - 0 \Bigg| \geq \frac{|a_1| + |a_2| + |a_3| + \dots |a_p|}{n} + \frac{|a_{p+1}| + |a_{p+2}| + \dots |a_n|}{n} < \dfrac{pk}n + \dfrac{\epsilon}{2n} (n-p) < \dfrac{pk}{n} + \dfrac{\epsilon}{2} \text{ for all } n \geq p.

As pp and kk are fixed, so limnpkn=0\displaystyle \lim_{n \to \infty} \dfrac{pk}{n} = 0

Corresponding to above ϵ>0\epsilon > 0, there exists a natural number qq such that

pkn<ϵ2 for all nq\dfrac{pk}{n} < \dfrac{\epsilon}{2} \text{ for all } n \geq q

In fact, here q=2pkϵ+1q = \lfloor \dfrac{2pk}{\epsilon} \rfloor + 1

So for nm= max(p, q)n \geq m = \text{ max(p, q)},

\[\Bigg| \dfrac{a_1 + a_2 + a_3 + \dots a_n}{n} \Bigg| < \dfrac{\epsilon}2 + \dfrac{\epsilon}2

\implies \displaystyle \lim_{n \to \infty} \frac{a_1 + a_2 + a_3 + \dots a_n}{n} = 0 \]

Corollary: If we take limnbn=l,0<l<\displaystyle \lim_{n \to \infty} b_n = l, 0 < |l| < \infty

Then taking xn=bnl,x_n = b_n - l, we have xnx_n \to \infty as nn \to \infty.

So limnx1+x2+x3+xnn=0.    limnb1+b2+b3+bnnl=0    limnb1+b2+b3+bnn=l\begin{aligned} \displaystyle \lim_{n \to \infty} \frac{x_1 + x_2 + x_3 + \dots x_n}{n} &= 0. \\ \implies \lim_{n \to \infty} \frac{b_1 + b_2 + b_3 + \dots b_n}{n} - l &= 0 \\ \implies \lim_{n \to \infty} \frac{b_1 + b_2 + b_3 + \dots b_n}{n} &= l \end{aligned}

Follow up problem

If xnn,xn>0{x_n}_n, x_n > 0 for all nn, converges to l,l>0,l, l > 0, then show that x1x2x3xnnl\sqrt[n]{x_1x_2x_3 \cdots x_n} \to l

We know that limnxn=l(xn>0,l>0)    limnlnxn=lnlUsing Cauchy’s First Limit Theoremlimnlnx1+lnx2++lnxnn=lnl    limnln(x1x2x3xnn)=lnl    x1x2x3xnnl as n\begin{aligned} \text{We know that } \displaystyle \lim_{n \to \infty} x_n &= l (x_n > 0, l > 0) \\ \implies \lim_{n \to \infty} \ln x_n &= \ln l \\ \text{Using Cauchy's First Limit Theorem} \\ \lim_{n \to \infty} \frac{\ln x_1 + \ln x_2 + \dots + \ln x_n}{n} &= \ln l \\ \implies \lim_{n \to \infty} \ln \left( \sqrt[n]{x_1x_2x_3 \cdots x_n} \right) &= \ln l \\ \implies \sqrt[n]{x_1x_2x_3 \cdots x_n} \to l \text{ as } n \to \infty \end{aligned}

Another follow up problem

Thanks to @Arghyadeep Chatterjee

Note by Adhiraj Dutta
1 year, 1 month ago

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Thanks a ton @Adhiraj Dutta . I have mentioned and linked your note in the problem I used this theorem. You can try the problem out here. I have deleted my non-Latex post on this theorem.

Arghyadeep Chatterjee - 1 year, 1 month ago

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You're welcome. Can you send the link to the question again?

Adhiraj Dutta - 1 year, 1 month ago

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Here is the link to the problem

Arghyadeep Chatterjee - 1 year, 1 month ago

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@Arghyadeep Chatterjee Thank you!

Adhiraj Dutta - 1 year, 1 month ago

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