Summation of $\frac{1}{k^2 - k}$ from $k=2$ to $\infty$.

Question

I couldn't get an idea how to get this summation?Can you help me please!!

Have you tried partial fractions? I don't know if it helps, but it may telescope. — pjs36, Jul 09 '15 at 02:43
Hint: There is a well known trick which starts by writing $\frac{1}{(k)(k-1)}$ as $\frac{1}{k-1}-\frac{1}{k}$ — lulu, Jul 09 '15 at 02:44
See here for ways to determine the partial sums (let $k=n+1$). Also, I can practically guarantee this is a duplicate. — Cameron Buie, Jul 09 '15 at 02:52
As for the downvote, I suspect it's because you've given no indication of your thoughts or attempts. Or perhaps because you seem to be yelling at us. — Cameron Buie, Jul 09 '15 at 02:53

score 3 · Accepted Answer · answered Jul 09 '15 at 02:44

3

Note that (for $k\geq 2$) $$ \frac{1}{k^2-k}=\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k} $$ so that $$ S_n\equiv\sum_{k=2}^n\frac{1}{k^2-k}=\sum_{k=2}^n\left(\frac{1}{k-1}-\frac{1}{k}\right)=\frac{1}{2-1}-\frac{1}{n}=1-\frac{1}{n}\cdot $$ What then can you say about $\lim_{n\to\infty}S_n$?

answered Jul 09 '15 at 02:44

Kim Jong Un

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Ok got it the answer is 1 !!Thnks!! – satyatech Jul 09 '15 at 02:50
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Isn't the telescope a wonderful invention. Thank you Hans Lippershey. – Mark Viola Jul 09 '15 at 02:52

Summation of $\frac{1}{k^2 - k}$ from $k=2$ to $\infty$.

1 Answers1