# convergence and limits

How can I rewrite $\frac{1}{1+nx}$ and prove it's absolute convergence as $n \rightarrow \infty$? Given $\epsilon > 0$, should I define $f_n(x)$ and $f(x)$? Any help is hugely appreciated. Thank you

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Convergence as $x\to$ something, as $n\to$ something? Uniform convergence? Pointwise convergence? – Pedro Tamaroff Mar 17 '13 at 20:25
@PeterTamaroff Apologies. I've edited the question. – Anona anon Mar 17 '13 at 20:27
also x cannot equal 0, obviously – Anona anon Mar 17 '13 at 20:30

It seems you're being given $$f_n(x)=\frac{1}{1+nx}$$ and asked to find its (pointwise) limit as $n \to \infty$. Note that if $x>0$, $\lim f_n(x)=0$. If $x=0$, $f_n(x)=1$ for each $n$, so $\lim f_n(x)=1$. There is a little problem when $x<0$, namely, when $x=-\frac 1 n$, so I guess we just avoid $x<0$. Thus, you can see for $x\geq 0$, you function is $$f(x)=\begin{cases}1\text{ ; if} x=0\\0\text{ ; if } x>0\end{cases}$$

In particular, convergence is not uniform over $[0,\infty)$ (since the limit function is not continuous), nor in $(0,\infty)$ because $$\sup_{(0,\infty)}|f_n-0| =1$$

for any $n$.

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That is my problem however - x cannot equal 0. How do I see if this sequence converges as n tends to infinity? – Anona anon Mar 17 '13 at 20:32
@Anonaanon Fix $x$ positive. Then $1+nx\to\infty$ as $n\to\infty$, yes? – Pedro Tamaroff Mar 17 '13 at 20:36
yep got it. Thank you so much Peter – Anona anon Mar 17 '13 at 21:10

First decide on what the answer will be. Let's not forget about $x=0$, the limit then is $1$. In all other cases, it is intuitively clear that the limit is $0$. If $x$ is negative, we may need to not start at $n=1$, in order to avoid division by $0$.

Now for the $\epsilon$ stuff. We look separately at the cases $x\gt 0$ and $x\lt 0$, and leave $x=0$ to you.

Fix $x\gt 0$. In order to make $\frac{1}{1+nx}\lt \epsilon$, it is enough to make sure that $1+nx\gt \frac{1}{\epsilon}$. This will certainly be true if $nx\gt \frac{1}{\epsilon}$, that is, if $n \gt \frac{1}{x\epsilon}$. So if for example we pick $N =\left\lfloor\frac{1}{x\epsilon}\right\rfloor$, then for every $n\gt N$ we will have $\frac{1}{1+nx}\lt \epsilon$.

Now we deal with the more challenging case where $x\lt 0$. Because negative numbers can cause confusion, replace $x$ by $-y$. So $y=|x|$. Then our general term is $\frac{1}{1-ny}$. If $ny\gt 1$, the absolute value of this is $\frac{1}{ny-1}$. We will want to make sure that $ny \gt 1$, so we will (among other things) ask that $n \gt\frac{1}{y}$.

We will also want $\frac{1}{ny-1}\lt \epsilon$. Sinc everything is positive, we can rewrite this as $ny \gt 1+\frac{1}{\epsilon}$, and then as $n\gt \frac{1}{y}+\frac{1}{y\epsilon}$.

Now let $N$ be the greater of $\left\lfloor \frac{1}{y}\right\rfloor$ and $\left\lfloor \frac{1}{y}+\frac{1}{y\epsilon}\right\rfloor$. If $n\gt N$, then $\left|\frac{1}{1+nx}\right|\lt \epsilon$.

Remark: We have given a very formal argument, supplying some intuition, but perhaps not enough. If some parts are a little too dense, they can be undensified.

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