The Leibniz rule says that if the alternating series is convergent the tail can be estimated by the magnitude of the first neglected term.

For instance in the series given above. If I wish to have an estimate of the tail.

I could bound it using the Leibniz rule it is:

meaning it is less than or equal to .166666666...

As I said in post #1 this is not the sharpest bound, that is what this thread is about.

]]>Can you please introduce yourself in Introductions.

]]>Can't see that ever changing...

Hard to predict the future.

Even this fellow could not:

]]>

This problem appears in another thread:

**A square with side A has two equilateral triangles inscribed within it also with side A. The triangles cross to form a rhombus. What is the area of the Rhombus in terms of A?**

**Also to clarify the equilateral triangles are on opposite sides of each other on the square. They both share the same length of A on the lines parallel from one another on the square.**

We will try to solve it using EM with the tools Geogebra and the Mathematica.

1) Open up Geo and make a slider called h and range it from 0 to 10 with an increment of 1. Set the animation to increasing and once.

2) Create points (0,0) and (h,0).

3) Use the regular polygon tool and click points A and B and press enter. A square will be created with sides of 1 unit.

4) Use the regular polygon tool and click A and then B and enter 3, a equilateral triangle ABE will be created.

5) Use the regular polygon tool and click C and then D and enter 3, a equilateral triangle DCF will be created.

6) Use the intersection tool and click on line segment FD and AE, point G will be created. Do the same for line segment FC and BE. Point H will be created.

7) Use the polygon tool and click on points F,G,E,H and back to F. Poly4 will be created. That is the rhombus we seek.

8) Bring up the spreadsheet.

9) Drag the slider h to 0.

10) Record poly4 to the spreadsheet.

11) Enter into the popup menu Row limit = 10 and uncheck show label and check trace to list.

12) Right click on h and check animation on. The spreadsheet will fill up with values. Copy the list in the top cell, you should get this:

{?, 0.154700538379252, 0.618802153517007, 1.39230484541327, 2.47520861406803, 3.86751345948130, 5.56921938165307, 7.58032638058334, 9.90083445627211, 12.5307436087194, 15.4700538379252, 15.4700538379252}

enter this into a notebook and manually, clean it up to look like this:

{0.154700538379252, 0.618802153517007, 1.39230484541327, 2.47520861406803, 3.86751345948130, 5.56921938165307, 7.58032638058334, 9.90083445627211, 12.5307436087194, 15.4700538379252, 15.4700538379252}

Wrap RootApproximant around it:

```
RootApproximant[{0.154700538379252, 0.618802153517007,
1.39230484541327, 2.47520861406803, 3.86751345948130,
5.56921938165307, 7.58032638058334, 9.90083445627211,
12.5307436087194, 15.4700538379252}]
```

we get,

Now what do we have here?

If we divide each element of the above list by the first element the pattern is uncovered.

```
FullSimplify[{1/3 (-3 + 2 Sqrt[3]),
1/3 (-12 + 8 Sqrt[3]), -9 + 6 Sqrt[3], 1/3 (-48 + 32 Sqrt[3]),
1/3 (-75 + 50 Sqrt[3]), -36 + 24 Sqrt[3], 1/3 (-147 + 98 Sqrt[3]),
1/3 (-192 + 128 Sqrt[3]), -81 + 54 Sqrt[3],
1/3 (-300 + 200 Sqrt[3])}/(1/3 (-3 + 2 Sqrt[3]))]
```

The output is:

{1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

therefore the relationship of the area of that rhombus R to the side A is

]]>Mathematica, the engine behind Wolfram also exhibits this behavior. For instance to follow bob bundy's example given in the other thread if we enter

we get

which is complex, instead of the desired result of 0. Entering

also yields

To get the answer of 0 which is the result Olinguito and Bob expect, we enter

this returns 0 which is as it should be.

]]>