The SummationSteps command accepts an expression that is expected to contain summations and displays the steps required to evaluate each summation given.

If expr is a string, then it is parsed into an expression using InertForm:-Parse so that no automatic simplifications are applied, and thus no steps are missed.  

The implicitmultiply option is only relevant when expr is a string.  This option is passed directly on to the InertForm:-Parse command and will cause things like 2x to be interpreted as 2*x, but also, xyz to be interpreted as x*y*z.

The output and displaystyle options are described in Student:-Basics:-OutputStepsRecord. The return value is controlled by the output option.  

This function is part of the Student:-Basics package.

$\mathrm{with}\left(\mathrm{Student}:-\mathrm{Basics}\right)\:$

$\mathrm{SummationSteps}\left(\mathrm{Sum}\left(i\,i=1..n\right)\right)$

$\begin{array}{lll}& & \sum _{i=1}^{n}i\\ \text{•}& & \text{Evaluate sum}\sum _{i=1}^{n}i\\ & & \frac{{\left(n+1\right)}^2}{2}-\frac{n}{2}-\frac{1}{2}\\ \text{•}& & \text{Simplify}\\ & & \frac{1}{2}{n}^2+\frac{1}{2}n\end{array}$

$\mathrm{SummationSteps}\left(1+2x+3\left(\mathrm{Sum}\left(2i\,i=1..n\right)\right)\right)$

$\begin{array}{lll}& & 1+2\cdot x+3\cdot \left(\sum _{i=1}^{n}2\cdot i\right)\\ \text{•}& & \text{Examine Sum}\\ & & \sum _{i=1}^{n}2\cdot i\\ \text{•}& & \text{Bring}2\text{outside of the sum}\\ & & 2\left(\sum _{i=1}^{n}i\right)\\ \text{•}& & \text{Evaluate sum}\sum _{i=1}^{n}i\\ & & 2\cdot \left(\frac{{\left(n+1\right)}^2}{2}-\frac{n}{2}-\frac{1}{2}\right)\\ \text{•}& & \text{Multiply}\\ & & {\left(n+1\right)}^2-n-1\\ \text{•}& & \text{Simplify}\\ & & n^2+n\\ \text{•}& & \text{This gives:}\\ & & 1+2\cdot x+3\cdot \left(n^2+n\right)\\ \text{•}& & \text{Simplify}\\ & & 3n^2+3n+2x+1\end{array}$

$\mathrm{SummationSteps}\left(\left(\mathrm{Sum}\left(i\,i=1..n\right)\right)+\mathrm{Sum}\left(j+2\,j=4..18\right)+\left(\mathrm{Sum}\left(4\left(k+1\right)\,k=10..n\right)\right)\right)$

$\begin{array}{lll}& & \left(\sum _{i=1}^{n}i\right)+\sum _{j=4}^{18}\left(j+2\right)+\sum _{k=10}^n\left(4\cdot k+4\right)\\ \text{•}& & \text{Examine Sum}\\ & & \sum _{i=1}^{n}i\\ \text{•}& & \text{Evaluate sum}\sum _{i=1}^{n}i\\ & & \frac{{\left(n+1\right)}^2}{2}-\frac{n}{2}-\frac{1}{2}\\ \text{•}& & \text{Simplify}\\ & & \frac{1}{2}{n}^2+\frac{1}{2}n\\ \text{•}& & \text{This gives:}\\ & & \frac{1}{2}{n}^2+\frac{1}{2}n+\sum _{j=4}^{18}\left(j+2\right)+\sum _{k=10}^n\left(4\cdot k+4\right)\\ \text{•}& & \text{Examine Sum}\\ & & \sum _{j=4}^{18}\left(j+2\right)\\ \text{•}& & \text{Reindex the summation so it starts at 1}\\ & & \sum _{j=1}^{15}\left(j+3+2\right)\\ \text{•}& & \text{Simplify and expand expression}\\ & & \sum _{j=1}^{15}\left(j+5\right)\\ \text{•}& & \text{Apply the sum rule:}\sum _{k=m}^n\left(\mathrm{a\_\_k}+\mathrm{b\_\_k}\right)=\left(\sum _{k=m}^n\mathrm{a\_\_k}\right)+\left(\sum _{k=m}^n\mathrm{b\_\_k}\right)\\ & & \left(\sum _{j=1}^{15}j\right)+\left(\sum _{j=1}^{15}5\right)\\ \text{•}& & \text{Evaluate the}\text{1st}\text{sum}\sum _{j=1}^{15}j\\ & & 120\\ \text{•}& & \text{Evaluate the}\text{2nd}\text{sum}\sum _{j=1}^{15}5\\ & & 75\\ \text{•}& & \text{Substitute evaluated sums into the expression}\\ & & 120+75\\ \text{•}& & \text{Simplify}\\ & & 195\\ \text{•}& & \text{This gives:}\\ & & \frac{1}{2}{n}^2+\frac{1}{2}n+195+\sum _{k=10}^n\left(4\cdot k+4\right)\\ \text{•}& & \text{Examine Sum}\\ & & \sum _{k=10}^n\left(4\cdot k+4\right)\\ \text{•}& & \text{Reindex the summation so it starts at 1}\\ & & \sum _{k=1}^{n-9}\left(4\cdot \left(k+9\right)+4\right)\\ \text{•}& & \text{Simplify and expand expression}\\ & & \sum _{k=1}^{n-9}\left(4k+40\right)\\ \text{•}& & \text{Apply the sum rule:}\sum _{k=m}^n\left(\mathrm{a\_\_k}+\mathrm{b\_\_k}\right)=\left(\sum _{k=m}^n\mathrm{a\_\_k}\right)+\left(\sum _{k=m}^n\mathrm{b\_\_k}\right)\\ & & \left(\sum _{k=1}^{n-9}4k\right)+\left(\sum _{k=1}^{n-9}40\right)\\ \text{•}& & \text{Bring}4\text{outside of the}\text{1st}\text{sum}\\ & & 4\left(\sum _{k=1}^{n-9}k\right)\\ \text{•}& & \text{Evaluate the}\text{1st}\text{sum}\sum _{k=1}^{n-9}k\\ & & 4\cdot \left(\frac{{\left(-8+n\right)}^2}{2}+4-\frac{n}{2}\right)\\ \text{•}& & \text{Multiply}\\ & & 2{\left(-8+n\right)}^2+16-2n\\ \text{•}& & \text{Evaluate the}\text{2nd}\text{sum}\sum _{k=1}^{n-9}40\\ & & 40n-360\\ \text{•}& & \text{Substitute evaluated sums into the expression}\\ & & 2{\left(-8+n\right)}^2+16-2n+\left(40n-360\right)\\ \text{•}& & \text{Simplify}\\ & & 2n^2+6n-216\\ \text{•}& & \text{This gives:}\\ & & \frac{1}{2}{n}^2+\frac{1}{2}n+195+\left(2n^2+6n-216\right)\\ \text{•}& & \text{Simplify}\\ & & \frac{5}{2}{n}^2+\frac{13}{2}n-21\end{array}$

$\mathrm{SummationSteps}\left(\mathrm{Sum}\left(\frac{1}{n!}\,n=1..\mathrm{\infty }\right)\right)$

$\begin{array}{lll}& & \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n!}\\ \text{•}& & \text{Apply the ratio test, which determines if the series diverges using}\underset{n\to \mathrm{\infty }}{lim}\left|\frac{a_{n+1}}{a_n}\right|=L\\ & & \left(\begin{array}{c}\text{If 0 ≤ L < 1 then}\sum _{n=0}^{\mathrm{\infty }}{a}_n\text{converges absolutely}\\ \text{If L > 1 then}\sum _{n=0}^{\mathrm{\infty }}{a}_n\text{diverges}\\ \text{If L = 1 then no conclusion is possible}\end{array}\right)\\ \text{•}& & \text{So we get}\\ & & \underset{n\to \mathrm{\infty }}{lim}\left|\frac{\frac{1}{\left(n+1\right)!}}{\frac{1}{n!}}\right|\\ \text{•}& & \text{Simplify inside expression}\\ & & \underset{n\to \mathrm{\infty }}{lim}\left|\frac{1}{n+1}\right|\\ \text{•}& & \text{Take the limit}\\ & & 0\\ \text{•}& & \text{Since the}0\le L\text{and}L<1\text{the infinte sum converges absolutely}\\ & & \left(\begin{array}{c}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n!}\text{converges}\end{array}\right)\end{array}$

The Student:-Basics:-SummationSteps command was introduced in Maple 2024.

For more information on Maple 2024 changes, see Updates in Maple 2024.