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For $ 0 \leq x < \infty$, I'd like your help with deciding whether the following series converges and uniformly converges: $\sum_{n=0}^{\infty}e^{-|x-n|}$.


I divided the sum into two sums where $N$ is the place where $x=n$, so the first sum is finite and the second one converges, so the total sum converges. Am I allowed to say that the sum is uniformly converges, since the second series, where the tail converges, does not depend on $x$?


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So is $x$ an integer, or something? – anon Feb 9 '12 at 9:55
It doesn't really say in the question,It's as I wrote it. I believe It's some kind of a series of function: $f_n(x)$. – Jozef Feb 9 '12 at 10:03
What you did shows that the series is pointwise convergent, but you can't conclude about uniform convergence since your $N$ depend on $x$. However, this can help you to show that the convergence is uniform on every bounded subset of $\mathbb R_+$. Then show that is the convergence is not uniform for all unbounded subset of $\mathbb R_+$. – Davide Giraudo Feb 9 '12 at 10:03
up vote 5 down vote accepted

Looking at the given series we see that uniform convergence is endangered by the fact that for arbitrary large $n$ we can find an $x$ where the $n$-th term is large, namely $\ =e^0=1$. Therefore we try to prove that the convergence is not uniform by exploiting this fact.

Put $s_n(x):=\sum_{k=0}^n e^{-|x-n|}$. By Cauchy's criterion uniform convergence means that for any $\epsilon>0$ there is an $n_0$, depending on $\epsilon$, such that $$\bigl|s_m(x)-s_n(x)\bigr|<\epsilon\qquad\forall m>n_0,\ \forall n>n_0,\ \forall x\in{\mathbb R}\ .$$ Assume now there is such an $n_0$ for $\epsilon:={1\over2}$ and put $$n:=n_0+1, \quad m:=n_0+2, \quad x:=m\ .$$ Then $s_n$ and $s_m$ differ by just one term (the $m$th), and one has $$\bigl|s_m(x)-s_n(x)\bigr|=e^{-|x-m|}=e^0>{1\over2}=\epsilon\ ,$$ which is a contradiction.

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Thanks! Why does $\bigl|s_m(x)-s_n(x)\bigr|=e^{-|x-m|}$? – Jozef Feb 9 '12 at 13:23
@Jozef: This is due to the fact that $n$ and $m$ differ by $1$. – JavaMan Feb 9 '12 at 18:52

For $0\leq x\in\mathbb{R}$, let q=$\lfloor{x}\rfloor\in\mathbb{N}$ be the greatest integer $\leq x$ and let $r=x-q\in[0,1)$ be the (nonnegative) fractional part of $x$, so that $x=q+r$. Then for $n\in\mathbb{N}$, $$ |x-n|= \left\{ \begin{matrix} (q-n)+r &\quad n\leq q\leq x \\ (n-q)-r &\quad n>x>q \end{matrix} \right. $$ so that $$ \begin{array}{} \sum_{n=0}^{\infty}e^{-|x-n|} &=& \sum_{n=0}^{q}e^{-r-(q-n)} &+& \sum_{n=q+1}^{\infty}e^{+r-(n-q)} \\&=& e^{-r}\sum_{n=0}^{q}e^{-(q-n)} &+& e^r\sum_{n=q+1}^{\infty}e^{-(n-q)} \\&=& e^{-r}\sum_{n=0}^{q}e^{-n} &+& e^{r-1}\sum_{n=0}^{\infty}e^{-n} \\&=& e^{-r}\frac{1-e^{-(q+1)}}{1-e^{-1}} &+& e^{r-1}\frac{1}{1-e^{-1}} \\&=& \frac{e^{r}+e^{1-r}+e^{-x}}{e-1} && \end{array} $$ converges absolutely (since each term is positive) -- but not uniformly! -- for all $x$.

(Aside:) If we extend the finite sum above (over $0\leq n\leq q$) to an infinite sum (over all $n\geq0$), we can derive a simple upper bound on the series of $\frac{2}{1-e^{-1}}$; in fact, the series sum always lies in $\left(\frac{2\sqrt e}{e-1},\frac{e+2}{e-1}\right]$.

Note, however, that in both cases, we have split the infinite sum on the left into two sums: a finite sum of increasing terms, stopping at $q$, the greatest integer less than or equal to $x$, and an infinite sum of decreasing terms. Both sums are geometric series and converge. The "cut" $q$, however, depends on $x$, and the largest term is either $e^{-(x-q)}=e^{-r}$ (when $r\leq\frac12$) or $e^{x-(q+1)}=e^{r-1}$ (when $r\geq\frac12$), which occurs twice in case $r=\frac12$. This largest term, then, depends on the distance $d(x)=\min(r,1-r)$ of $x$ from $\mathbb{Z}$ in $\mathbb{R}$, which always lies in $[0,\frac12)$, and so has value $e^{-d(x)}$ which, in particular, is always $\geq e^{-\frac12}$.

Consequently, the series is not uniformly convergent. According to the definition of uniform convergence (of a sequence of functions $f_n$ from some set to $\mathbb{R}$ or $\mathbb{C}$ or a metric space):

A sequence of functions $f_n:X\rightarrow\mathbb{R}$ is uniformly convergent if, firstly:

for each $x\in X$, $f_n(x)$ converges pointwise to a finite value $f(x)$, so that there exists a well-defined limit function $f:X\rightarrow\mathbb{R}$, in which case we say that $(f_n)$ converges and write $f_n\rightarrow f$,

and secondly, if:

for each $\epsilon>0$, there exists an $N\in\mathbb{N}$ so that for all $n \geq N$ (and all $x\in X$), we have $|f_n(x)-|f(x)|<\epsilon$.

In particular, $N=N(\epsilon)$ can depend on $\epsilon$ but not on $x\in X$. This is the sense of the word uniform.

To apply this definition to our series, we need to find the sequence of functions to which the definition applies. This will be the sequence $f_n$ of partial sums $$ f_n(x)=\sum_{k=0}^{N}e^{-|x-k|} $$ and to prove its uniform convergence, we would want to show that the differences $$ |f_n(x)-f(x)|=\sum_{k=N+1}^{\infty}e^{-|x-k|} $$ could be made smaller than any desired $\epsilon>0$, for all $x$, given a suitable choice of (just one value of) $N$. In our case, we can't provide an $N$ ahead of time, without foreknowledge of $x$, since the terms peak near $x$, as discussed above.

An equivalent formulation for uniform convergence is that the sequence $\alpha_n=\sup_{x\in X}|f_n(x)-f(x)|$ tends toward zero as $n\rightarrow\infty$.

However in our case, using $x=n+1$ to bound $\alpha_n$ below, $$ \alpha_n =\sup_{x\geq0}\sum_{k=n+1}^{\infty}e^{-|x-k|} \geq\frac1{1-e^{-1}}>0 \quad\implies\quad \alpha_n\nrightarrow 0 \quad\text{as}\quad n\rightarrow\infty $$ showing that the convergence is not uniform. (Aside: In fact, $\alpha_n=\frac{1+e^{-1}}{1-e^{-1}}=\frac{e+1}{e-1}$.)

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How can you find a bounded r when n <= x and n goes to infinity. – citrucel Feb 9 '12 at 13:16
Since $q$ depends on $x$, this does not prove uniform convergence (and in fact, there is no uniform convergence). – Did Feb 9 '12 at 16:52
This is irrelevant. Please check the definition of uniform convergence or use the lemma: if $\sum f_n$ converges uniformly, then $\|f_n\|_\infty\to0$. – Did Feb 9 '12 at 18:43
@DidierPiau: thanks, I think it's corrected now. – bgins Feb 10 '12 at 0:30

We can show that the series is uniformly convergent on bounded subsets of $\mathbb R_+$. Indeed, let $B$ such a set and $R$ an integer such that $B\subset [0,R]$. If $n>R$ then $n-x>R-x\geq 0$ os $|n-x|=n-x$ and $0\leq e^{-|n-x|}=e^{-(n-x)}=e^{-n}e^x\leq e^{-n}e^R$, so in fact the series $\sum_{n\geq 0}e^{-|n-x|}$ is normally convergent on $[0,R]$ hence on $B$.

If $U$ is an unbounded subsets of $\mathbb R_+$, put $I_k:=[k,k+1[$. Let $n_k$ an increasing sequence of integers such that $I_{n_k}\cap U\neq\emptyset$. Then for each $n\in \mathbb N$, and $k'$ such that $n< n_{k'}$ $$\sup_{x\in U}\sum_{k=n}^{+\infty}e^{-|k-x|}\geq \sup_{x\in U}\sum_{j=k'}^{+\infty}e^{-|n_j-x|}\geq \sup_{x\in U\cap I_{n_{k'}}}e^{-|n_{k'}-x|},$$ and for $x\in U\cap I_{n_{k'}}$ we have $-|n_{k'}-x|=-x+n_{k'}\geq -1$ so $\sup_{x\in U}\sum_{k=n}^{+\infty}e^{-|k-x|}\geq e^{-1}$ which shows that the convergence cannot be uniform on $U$.

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