Seriesly
Prove that k.k! = (k+1)! - k! and sum the series 1.1! + 2.2! + 3.3! +...+n.n!
Problem
Prove that
$$k \times k! = (k+1)! - k!$$ and sum the series
$$1 \times 1! + 2 \times 2! + 3 \times 3! +...+n \times n!$$
A telecoping series is a series that can be written as the difference of two expressions in such a way that almost all the terms cancel with the following or preceding term leaving a few terms which can be combined to give the sum of the series. See the Wikipedia article Telescoping Series.
Getting Started
When proving an identity it is usually best to start with the more complicated expression and simplify it so this time start with the expression on the right hand side $(k+1)! - k!$.
Student Solutions
This task previously appeared on the main NRICH site, with solutions as follows:
Thank you for these solutions to Shahnawaz Abdullah; Daniel, Liceo Scientifico Copernico, Torino, Italy; Anderthan, Saratoga High School; Andrei, School 205, Bucharest, Romania; David, Queen Mary's Grammar School, Walsall; Paddy, Peter, Greshams School, Holt, Norfolk; Ngoc Tran, Nguyen Truong To High School (Vietnam); Chris, St. Bees School; Dorothy, Madras College; A Ji and Hyeyoun, St. Paul's Girls' School; and Yatir from Israel.
To prove that $k \times k! = (k+1)! - k!$. If we take $k!$ out as a factor from the right hand side of the equation, we are left with $k! \times ((k+1)-1)$ which simplifies to $k \times k!$, as required.
Now we sum the series $1 \times 1!+.....n \times n!$ As we have proved, $n \times n!$ is equal to $(n+1)! - n!$ and therefore $(n-1) \times (n-1)!$ is equal to $(n-1+1)! - (n-1)!$ which simplifies to $n! - (n-1)!$.
If we add the two results, we find that $n!$ cancels. If we sum the series from 1 to $n$, we find that all of the terms cancel except for $(n+1)!$ and $-(1!)$. Thus the sum of all numbers of the form $r \times r!$ from $1$ to $n$ is equal to $(n+1)! - 1$.
Teachers' Resources
Why do this problem?
This problem is an example of a "method of differences" or "telescoping series" problem which does not involve partial fractions. Instead, it provides practice in manipulating factorial expressions and results in a surprisingly neat sum for a series which grows very rapidly.
Possible approach
This problem could be used as a starter or extra challenge for students well-versed in factorials and familar with the the method of differences. If factorials are less familar, you could start with some basic practice in manipulating algebraic expressions involving factorials, e.g. $\frac{(k+1)!}{(k-1)!}$, $(k+1)\times k!$, etc.
This priming should be sufficient for students to see the factorisation required to prove the initial identity. If not, the further hint from the "Getting Started" section could be given (When proving an identity it is usually best to start with the more complicated expression and simplify it so this time start with the expression on the right hand side $(k+1)! - k!$. )
The fact that proving this identity is the first stage of the problem should again be sufficient for students to realise that the identity can be used to re-write the terms of the series they are required to sum. For students who have never met the method of differences, it may be helpful to suggest that they write out the first three and last three terms of the series in this expanded form. They may also benefit from writing each term vertically below the previous (with dots in the middle to represent the many unspecified terms following the same pattern). This should help them to see the cancellation of terms and reach a concise expression for the sum.
Key questions
What is the highest common factor of $(k+1)!$ and $k!$?
Which part of each term cancels with part of a subsequent term? What remains after all cancellations have been removed?
Possible support
Students could work on both parts of the problem numerically for different small values of $k$ and $n$, for example showing that $4! - 3! = 3!\times (4-1) = 3\times 3!$ or that $(2! - 1!) + (3! - 2!) + (4! - 3!) = 4! - 1$.