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Let $a<b$, $a,b\in R$ Prove that the series $\sum_{n=1}^\infty \frac{1}{n(n-x)}$ converges absolutely and uniformly on [a,b]\N

What is [a,b]\N? I am really confused. For absolute convergence I know we take the absolute of the series and for uniform we need Weistress or Abel's uniform convergence test.

Any help much appreciated!

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    $\begingroup$ "for uniform we need Weistress or Abel's uniform convergence test" No, you don't know that. What you know is that these tests are often useful. $\endgroup$ – zhw. Jan 21 '16 at 2:47
  • $\begingroup$ ok we can do it by finding the supremum? $\endgroup$ – cab Jan 21 '16 at 2:51
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$[a,b]\setminus\mathbb N$ is a finite closed interval in $\mathbb R$ in which any non-negative integers are removed. For example, $[0,\pi]\setminus\mathbb N=(0,1)\cup(1,2)\cup(2,3)\cup(3,\pi]$.

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  • $\begingroup$ ah ok thanks, so in this case its the reals less non negative integers. $\endgroup$ – cab Jan 21 '16 at 2:00
  • $\begingroup$ Well not quite, $[a,b]\setminus\mathbb N$ won't contain any reals greater that $b$ for example. You're not given one specific domain. Rather, a type of domain. $\endgroup$ – Tim Raczkowski Jan 21 '16 at 2:02
  • $\begingroup$ ah yes of course. So how does one attempt such a question? $\endgroup$ – cab Jan 21 '16 at 2:03
  • $\begingroup$ Absolute convergence can be shown by a straightforward application of the limit comparison test with $\sum 1/n^2$. $\endgroup$ – Tim Raczkowski Jan 21 '16 at 2:11
  • $\begingroup$ it's that simple? I don't understand the effect of [a,b]\N? Because that solution is the same as a normal solution. Also x is just a value, right? $\endgroup$ – cab Jan 21 '16 at 2:17
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Fact: Suppose $f_n$ is a sequence of real-valued functions defined on a set $E.$ Let $N\in \mathbb N.$ If $\sum_{n=N}^{\infty} f_n$ converges uniformly on $E,$ then $\sum_{n=1}^{\infty} f_n$ converges uniformly on $E.$ I'll use this below.

In our problem, it's enough to show that the series converges uniformly on $[-N,N]\setminus \mathbb N$ for each $N \in \mathbb N.$ Fix such an $N.$ Then for $n>N,$ we have

$$\left |\frac{1}{n(n-x)} \right| \le \frac{1}{n(n-N)}$$

for all $x \in [-N,N].$ Because $\sum_{n=N+1}^{\infty} 1/n(n-N) < \infty,$ we see $ \sum_{n=N+1}^{\infty} 1/n(n-x)$ converges uniformly on $[-N,N]$ by the Weierstrass M-test. This implies $ \sum_{n=1}^{\infty} 1/n(n-x)$ converges uniformly on $[-N,N]\setminus \mathbb N$ as desired.

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