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So we learned telescoping series in class and I came across this question in my textbook and I tried to evaluate it, but I don't understand how to do it.

$$ \sum\limits_{k=1}^n\left(k \cdot k!\right) $$

According to the answers:

$$ \sum\limits_{k=1}^n\left(k \cdot k!\right) = (n+1)! - 1 $$

We learned how to simplify certain telescoping series using partial fractions but I don't have any idea on how to start this question.

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Note that $$k \cdot k! =(k+1)! - k!$$ Now telescoping should help you.

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    $\begingroup$ Indeed. These kinds of telescoping series problems will almost always become trivial if you determine a way of rewriting the summand as a difference of "similar" expressions. $\endgroup$ – oldrinb May 5 '13 at 4:04
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    $\begingroup$ Thank you very much! Now the question is much more simple. I completely agree with you oldrinb, finding the "right" expression is usually the most difficult part. $\endgroup$ – user2351149 May 5 '13 at 4:16
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    $\begingroup$ @user2351149 A simple way to get the right expression when you know what the answer is, is as follows. If you want to show $$\sum_{k=1}^n f(k) = S(n)$$ try to see if $f(k) = S(k+1) - S(k)$. In this case, $((n+1)!-1) - (n!-1) = n \cdot n!$. $\endgroup$ – user17762 May 5 '13 at 4:20

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