Example 26.1.1
Examine for which values of r the geometric series:      converges.

In order to see under which circumstances this series converges we form the n-th partial sum and take the limit of this partial sum for  -> .

>	s[n]:=Sum(r^k, k=0..n);

Warning:   s[n]  is an element of the Maple table with alphanumeric index  n.

                    s[n] is NOT a function of  n  in Maple sense !!!

                    Having used  s[n]  for the n-th partial sum, we will use  S  for the infinite sum.

>	S:=Limit(s[n], n=infinity);

>	S:=value(S);

>	S:=simplify(S);

>

Clearly this limit exist if and only if  -1 < r < 1.  

Maple 9 has a problem with this limit (even when the assume command is used to tell the system that -1 < r < 1).  Sorry!  we have to wait for the next release for this bug to be fixed.

Example 26.1.2
Does the series  converge?  Explain your answer.

>	S:=Sum((2/Pi)^n, n=1..infinity);

This is a geometric series and according to the above it will converge because  -1 <  < 1.

>	r:=evalf(2/Pi);

>	S:=value(S);

>

Example 26.1.3
Does the series    converge?

We will once again investigate the general term of the sequence.

>	S:=Sum(2^(6*n+5)/5^(2*n-7), n=1..infinity);

>	a[n]:=op(1, S);

>	simplify(evalf(a[n]));

Clearly this geometric series diverges because  > 1.

>	S:=value(S);

>

Example 26.1.4
Determine if the series    converges.

Certain series converge because the terms or pieces thereoff cancel against each other.  The easiest way of discovering such behaviour is by applying partial fraction decomposition to the individual terms of the sequence.

>	S:=Sum(5/(n*(n+1)), n=1..infinity);

>	s[n]:=Sum(5/(k*(k+1)), k=1..n);

>	a[n]:=op(1, S);

>	a[n]:=convert(a[n], parfrac, n);

For clarity we display six successive terms.

>	for m from 0 to 5 do

>	a[n+m]:=subs(n=n+m, a[n]);

>

od;

If we substitute  n = 1  in the above and sum the terms, then successive items will cancel and we obtain:  =  =  =  .

>	s[6]:=value(subs(n=6, s[n]));

Similarly:    + ... +  .

>	s[n]:=value(s[n]);

Of course S must equal .   Maple will readily acknowledge that identity.

>	evalb(limit(s[n], n=infinity)=value(S));

>

Example 26.1.5
Show that     diverges.

>	S:=Sum((n^2+5)/(3*n^2+11), n=1..infinity);

>	a[n]:=op(1, S);

>	L:=limit(a[n], n=infinity);

This means of course that  diverges, which is easily verified by the system.

>	S:=value(S);

>