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I have been trying to prove the uniform convergence of a sequence of functions defined by $f_n(x) = e^{(n+1)x/n}$ by the epsilon definition of uniform convergence.

I have found the pointwise limit of the sequence, $f(x)=e^x$. I am having trouble picking $n>N$ such that for all $\epsilon > 0$, $|f_n(x)-f(x)|< \epsilon$ showing that $f_n$ converges uniformly on $[0, 5]$.

Please help!

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    $\begingroup$ Hi @SaagarS! It will help readers if you use MathJax to render your math. $\endgroup$ Jul 28 '20 at 21:07
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    $\begingroup$ Hmm. Do you really need $N$ to be independent of $\varepsilon$? You're also missing a quantifier on $x$ in your request. $\endgroup$ Jul 28 '20 at 21:10
  • $\begingroup$ Hi Adina, thanks for your help with MathJax. I edited the request accordingly! $\endgroup$
    – SaagarS
    Jul 28 '20 at 21:15
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Using necessary and sufficient condition for uniform convergence (here), we have: $$\lim_{n \to \infty}\sup_{x \in [0,5]}\left| e^{(n+1)x/n} - e^x\right| =\lim_{n \to \infty}\sup_{x \in [0,5]} e^5\left| e^{x/n} - 1\right| = 0$$

If it is so necessary, then $N$ you can obtain from $e^5\left| e^{5/n} - 1\right| \leqslant 1$.

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So, with uniform convergence, by definition, you should be solving the following problem.

Given $\epsilon > 0$, pick $N \in \mathbb{N}$ such that for all $n\geq N$ and for all $x\in[0,5]$, $$|f_n(x) - f(x)| < \epsilon.$$

Note that this definition allows $N$ to depend on $\epsilon$, but not on $x$.

Often with increasing or decreasing functions, it is useful to bound the function with its value at one endpoint of the given interval.


Here is a full solution. For $x\geq 0, \alpha \geq 1$, $e^{\alpha x}$ is greater than or equal to $e^x$, because the exponential function is increasing. So first off, we can drop the absolute value. \begin{align} |f_n(x) - f(x)| &= e^{\frac{(n+1)}{n}x} - e^x\\ &= e^{x + \frac{1}{n}x} - e^x\\ &= e^x(e^\frac{x}{n} - 1)\\ \end{align}

This last expression is a product of two positive increasing functions of $x$, so is largest when $x$ is largest. So we can bound it on the interval $[0,5]$ as follows:

\begin{align} e^x(e^\frac{x}{n} - 1) &\leq e^5(e^\frac{5}{n}-1)\\ \end{align}

Now, we are interested in $n \geq N$ for some fixed choice of $N$. This means $e^\frac{5}{n} \leq e^\frac{5}{N}$. So, to recap, we want to pick $N$ so that the following is less than $\epsilon$:

$$ |f_n(x) - f(x)| \leq e^5(e^\frac{5}{n}-1) \leq e^5(e^\frac{5}{N}-1) < \epsilon$$

Now what remains is to rearrange the last inequality to decide how to pick $N$, based on $\epsilon$. After rearranging, we get:

\begin{align} &e^\frac{5}{N} < \epsilon e^{-5} + 1\\ \iff &\frac{5}{N} < \ln (\epsilon e^{-5} + 1)\\ \iff &N > \frac{5}{\ln (\epsilon e^{-5} + 1)} \end{align}

To make $N$ a large enough integer, it is enough to round up the expression on the right hand side to the nearest integer, and choose that for $N$.

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  • $\begingroup$ Thank you for this. My problem is with picking N! $\endgroup$
    – SaagarS
    Jul 29 '20 at 6:02
  • $\begingroup$ @SaagarS I have added a full solution to help you out! $\endgroup$ Jul 29 '20 at 15:09
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If you have $f_n(x)=f(x+x/n)$ x$\in$ [0,5] and f is uniformly continuous function on [0,10] its a good exercise to show that $f_n->f$ uniformly.Here you have $f(x)=e^x$ which is Lip-continuous on [0,10] so its uniformly continuous.In mathematics it's really important if you choose to give some general-difficult definitions for your theory to come up with good results of your definitions(theorems that come straight from your definitions)or you can use easy definitions and then results coming from your definitions are much more difficult!

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