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What exactly converges the series?

$\sum _{k=3}^{\infty \:}\frac{2}{k^2+2k}$

I tried taking out the constant $=2\sum _{k=3}^{\infty \:}\frac{1}{k^2}$

then $p=2,\:\quad \:p>1\quad \Rightarrow \sum _{k=3}^{\infty \:}\frac{1}{k^2}$

but I really don't know what i'm doing

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up vote 3 down vote accepted

Use partial fractions: Suppose



All the terms cancel out except four:

$$f(n)=\displaystyle \frac{1}{3}+\frac{1}{4}-\frac{1}{n+1}-\frac{1}{n+2}=\frac{7}{12}-\frac{2n+3}{(n+1)(n+2)}$$

Obviously the last fraction converges to zero and you 're left with $\frac{7}{12}$.

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Note that $\dfrac{2}{k^2+2k}=\dfrac{1}{k}-\dfrac{1}{k+2}$. So our sum is $$\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{1}{8}\right)+ \left(\frac{1}{7}-\frac{1}{9}\right)+\left(\frac{1}{8}-\frac{1}{10}\right)+\cdots.$$Remove the parentheses, and note the mass cancellations (telescoping). Almost everything dies, and we are left with $\dfrac{1}{3}+\dfrac{1}{4}$.

Remark: The above argument was informal. It can be replaced by a formal argument in which we take the sum $s_n$ of the first $n$ terms, and show that $\lim_{n\to\infty}s_n=\frac{1}{3}+\frac{1}{4}$.

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Hint: $ \dfrac{2}{k^2+2k}=\dfrac{(k+2)-k}{k(k+2)}=\dfrac{1}{k}-\dfrac{1}{k+2}$

So, $\sum\limits_{k=3}^\infty \dfrac{2}{k^2+2k}=\bigg( \dfrac13-\dfrac15\bigg)+\bigg(\dfrac14-\dfrac16\bigg)+\bigg(\dfrac15-\dfrac17\bigg)+\ldots$

Does this help?

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You should only sum up to $N$, not $\infty$. As it stands now, you're writing a convergent sum as the different of two divergent sums, which is not correct. – Hans Lundmark Apr 3 '14 at 7:35
@mrf Is there anyway to discard my edits? – Rustyn Apr 3 '14 at 7:39
@Rustyn I rolled it back for you. – mrf Apr 3 '14 at 7:40
@mrf thanks, I appreciate it. – Rustyn Apr 3 '14 at 7:41
@mrf is the post okay now? – Hawk Apr 3 '14 at 7:41

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