I am aware that this has a duplicate but I am trying to prove it differently than others.

Proposition: If $\lim_{n\to\infty} a_n = L \neq 0,$ then $\lim_{n\to\infty} \frac{1}{a_n} = \frac{1}{L}$


$\lim_{n\to\infty} a_n = L$, so $\exists N_2 $st $\forall n > N_2 |a_n-L|< \epsilon$

Take N = max$(N_1,N_2); $ so $\forall n>N, |a_n-L| < \epsilon $

and $\forall n>N_1, |a_n| < \frac{1}{|L|}$ (by Lemma)

So $|\frac{1}{a_n}-\frac{1}{L}| = \frac{|a_n-L|}{|a_n|*|L|} = |a_n-L|*\frac{1}{L}*\frac{1}{a_n}$

$<\epsilon *\frac{1}{L}*L$


I am just having difficulty proving the Lemma, which should be $|a_n|<\frac{1}{|L|}$

Any advice?

  • $\begingroup$ variable is $n$ not $x$ $\endgroup$ – Chinnapparaj R Apr 24 '19 at 15:19
  • $\begingroup$ Sorry, I copy pasted the format for fractions. Is there any other advice you can give ? $\endgroup$ – beepbeepboop123123 Apr 24 '19 at 16:50
  • 1
    $\begingroup$ @beepbeepboop123123 Your question had a tag of just "elementary-number-theory". However, what you're asking doesn't really fit as the tag description says "Questions on congruences, linear Diophantine equations, greatest common divisor, divisibility, etc.". Instead, a tag like "limits" would be more appropriate, so I've taken the liberty of changing your tag. You may wish to check Tags for any other appropriate tags to add to your post. $\endgroup$ – John Omielan Apr 24 '19 at 23:50

Your proof attempt has a few issues. For one, it changes the limit from $A$ to $L$ (I will use $L$ here). Also, it doesn't define what $N_1$ is, but from the context it seems to be for the minimum value where $\left|a_n - \frac{1}{L}\right| \lt \epsilon$. Finally, and most importantly, you can't prove your suggested lemma of $\left| a_n \right| \lt \frac{1}{\left| L \right|}$ because it's not necessarily true. For example, if $a_n = 2 + \frac{1}{n}$, then $L = 2$, but you never have $\left|a_n\right| \lt \frac{1}{2}$.

The way I would do it instead is to expand the part in absolute values and then use the resulting inequalities to place an appropriate upper limit on $\frac{1}{\left| a_n \right|}$. In particular, you have

$$\left| a_n - L \right| \lt \epsilon \; \Rightarrow \; -\epsilon \lt a_n - L \lt \epsilon \; \Rightarrow \; -\epsilon + L \lt a_n \lt \epsilon + L \tag{1}\label{eq1}$$

There are $2$ main cases to consider, i.e., $L$ being positive or negative. I'll show how to handle the positive case. First, have $\epsilon \lt \frac{L}{2}$. As such, all $3$ parts of the end of \eqref{eq1} are positive. Thus, multiplying the left & middle parts by $\frac{1}{a_n\left(L - \epsilon\right)}$ gives

$$\frac{1}{a_n} \lt \frac{1}{L - \epsilon} \lt \frac{2}{L} \tag{2}\label{eq2}$$

with the last inequality coming from $L - \epsilon \gt \frac{L}{2}$ by the earlier assumption. This gives that

$$\left|a_n - L\right| \times \frac{1}{L} \times \frac{1}{\left|a_n\right|} \lt \frac{2\epsilon}{L^2} \tag{3}\label{eq3}$$

The epsilon to choose for proving the limit of $\frac{1}{a_n}$, call it $\epsilon_2$, could be any $\epsilon_2 = \frac{2\epsilon}{L^2} \lt L$. In that case, the original epsilon would be $\epsilon = \frac{L^2\epsilon_2}{2}$ to give an appropriate $N$ to be used.

Handling the case where $L$ is negative is similar, except you need to use the right hand part of \eqref{eq1} instead and make sure to use absolute values in appropriate places. I'll leave this part to you to finish.

  • $\begingroup$ Thank you very much for the advice, but I have a couple of questions; 1) After multiplying the left and middle parts by $\frac{1}{a_n(L-\epsilon}$ how did you get $\frac{2}{L}$ on the right hand side of (2)? 2) When did we discover $L-\epsilon>\frac{L}{2}$? 3) How did you realize to set $\epsilon<\frac{L}{2}$? $\endgroup$ – beepbeepboop123123 Apr 25 '19 at 14:29
  • $\begingroup$ @beepbeepboop123123 You're welcome. As for (1), if $a$ and $b$ are positive quantities with $a \gt b$, then multiplying by $\frac{1}{ab}$ gives $\frac{1}{b} \gt \frac{1}{a}$. In this case, $L - \epsilon \gt \frac{L}{2} \; \Rightarrow \; \frac{2}{L} \gt \frac{1}{L - \epsilon}$. For (2), since I chose $\epsilon \lt \frac{L}{2}$, then $-\epsilon \gt -\frac{L}{2} \; \Rightarrow \; L - \epsilon \gt \frac{L}{2}$. For (3), I needed to set a reasonable upper bound on $\epsilon$ to get a good upper bound in my eq. (2). $\frac{L}{2}$ works, but I could've used others, like $\frac{L}{3}$, instead. $\endgroup$ – John Omielan Apr 25 '19 at 15:08

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