3
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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

$\endgroup$
3
  • 3
    $\begingroup$ 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. $\endgroup$ – obataku May 5 '13 at 4:04
  • 1
    $\begingroup$ 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
  • 3
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.