Here is a fun video about this.

]]>I do not understand what you want.

]]>```
SequenceLimit[{-0.5`, -0.5833333333333334`, -0.6345238095238095`, \
-0.6628718503718504`, -0.6777662022075269`, -0.685395708269249`, \
-0.6892561888834033`, -0.691197870228108`, -0.6921715717324427`, \
-0.6926591377284108`}]
```

The amount of math and other things stuffed into mathematica is unbelievable.

]]>Good to see you getting back into math and smart to use some modern tools like Geogebra to assist you.

]]>This problem appears in another thread and someone asked to see my solution.

1) Find a fifth degree polynomial f(x) with the conditions (x-1)^3 | f(x) -1 and x^3 | f(x).

The outline of how it is done.

We get the remainders:

Those remainders on the RHS must be set to 0 and the coeficients equated to 0.

From that we get 6 simultaneous equations that are easy to solve:

So the polynomial we seek is

Of course it is much easier using Mathematica:

Way 1:

```
p[x_, a_, b_, c_, d_, e_, f_] := a*x^5 + b*x^4 + c*x^3 + d*x^2 + e*x + f
rul = FindInstance[
ForAll[x,
PolynomialRemainder[p[x, a, b, c, d, e, f] - 1, (x - 1)^3, x] ==
0 && PolynomialRemainder[p[x, a, b, c, d, e, f], x^3, x] ==
0], {a, b, c, d, e, f}] // First;
p[x, a, b, c, d, e, f] /. rul // TraditionalForm
```

yields:

6 x^5-15 x^4+10 x^3

Way 2:

```
p[x_, a_, b_, c_, d_, e_, f_] :=
a*x^5 + b*x^4 + c*x^3 + d*x^2 + e*x + f;
PolynomialRemainder[p[x, a, b, c, d, e, f] - 1, (x - 1)^3, x];
PolynomialRemainder[p[x, a, b, c, d, e, f], (x)^3, x];
rul = Solve[{-1 + 6 a + 3 b + c + f == 0, (-15 a - 8 b - 3 c + e) ==
0, (10 a + 6 b + 3 c + d) == 0, f == 0, e == 0, d == 0}, {a, b,
c, d, e, f}][[1]];
p[x, a, b, c, d, e, f] /. rul
```

10 x^3 - 15 x^4 + 6 x^5

]]>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.

]]>