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Show that $$\sum_{n=2}^\infty \frac{(-1)^n}{n^2-n}\sim 2 \ln(2)-1$$ and $$\sum_{n=2}^\infty \frac{1}{n^2-n}\sim 1$$

How do I approach this problem?

I know the first one is an alternating series where $\sum_{n=2}^\infty \frac{(-1)^n}{n-1}\sim \ln(2)$.. But so what?

The second one, I could take the long road and do: $$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\cdots=1$$

But that would take a very long time.

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  • $\begingroup$ Is there a reason why you write $\sim$ instead of $=$? $\endgroup$ – OneAndOnlyDaniel Jun 2 '19 at 16:42
  • $\begingroup$ @OneAndOnlyDaniel Just following the notation used in my curriculum book, as the sum of a series will not always equal the exact number A. $\endgroup$ – mahma Jun 2 '19 at 16:54
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First one: apply partial fractions to get,

$$\sum_{n=2}^\infty (-1)^n * (\frac{1}{n-1} - \frac {1}{n}) = \ln(2) + \ln(2)-1=2\ln(2)-1$$

Why do you think that method for 2nd one is long?

$$S= \frac 1 {1*2 } + \frac 1 {2*3} + \frac 1 {3*4}+.... $$

$$ = (\frac 1 1 - \frac 1 2) + (\frac 1 2 - \frac 1 3) + (\frac 1 3 - \frac 1 4)+ \ldots $$

$$=1 $$

(again by partial fractions)

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