@Keshav4  If a> = 0 then obviously your function  f  is increasing and it takes its smallest value at  x = 0 . For negative values of the parameter  a , the example can be solved only numerically.

@Prakash J 

plot(abs@h, 1..2, labels=["t","h(t)"]);

@Scot Gould  The disadvantage is that the syntax is longer, and you yourself have already written about the advantages. I usually don't use this option because it is quite possible to do without it.

@vv  and @Carl Love  Thank you for answers.

You can arrange this calculation without assignments and conversions:

restart; 
Sum(GAMMA(2*b-1+2*n)*GAMMA(2*n-s)*(b-1/2+2*n)/(GAMMA(2*b+2*n+s)*GAMMA(2*n+1)), n = 0 .. infinity); 
subs(Sum = sum, %) = eval(%, Sum = sum);

@Jean-Michel Collard   May be 

https://www.mapleprimes.com/questions/231554-10--The-New-Way-To-Calculate

@nm  Omega &x v(t)  means the cross product of the vectors  Omega  and  v(t)  in LinearAlgebra package.

Upload your worksheet (or a link to it) here for localizing the problem (using the bold green arrow in the MaplePrimes editor). As a last resort, just copy and paste here the code in which this error occurs.

@minhthien2016  Geometrically, all the examples vv provided differ little from each other. All of these tetrahedrons are similar to your original example  (0, 0, 0), (3, 0, 0), (0, 5, 0), (0, 0, 14) . They are obtained from your original example by shifting by some vector, stretching and rotating in space. The easiest way to see this is by plotting. The code below plots your example and the first example from vv's code:

restart;
A:=plottools:-tetrahedron([[0,0,0],[3,0,0],[0,5,0],[0,0,14]]):
B:=plottools:-tetrahedron([[-9,1,-7],[-18,-17,59],[-79,91,8],[243,183,77]]):
plots:-display(< A | B >, scaling=constrained);

 
Below is an example of a tetrahedron that is not in my previous answer. Its inscribed sphere has  unit radius and its center is [4, 2, 1] :

A:=plottools:-tetrahedron([[0,0,0],[5,0,0],[5,5,0],[5,3,4]]):
B:=plottools:-sphere([4, 2, 1], 1):
plots:-display(A,B, scaling=constrained, transparency=0.3, axes=normal, view=[0..5.5,0..5.5,0..4.3]);

I don't know of any other way to get substantially different examples than "brute force".

@mehran rajabi 

restart;
asympt(1/(x^2-1), x, 10);

# Or
Sum(1/x^(2*n), n=1..infinity);

# Or automatically
convert(1/(x^2-1), FormalPowerSeries, x=infinity);

 

@vv  Thanks for the helpful comment. I tweaked the code a bit. Now it immediately returns the exact single solution.

@Carl Love  You wrote "There are no real roots". If  fsolve  cannot find a real root, this does not mean that there are no real roots.

Example (in Maple 2018):

fsolve(x^1000-1000^x-1, x=999..1001);
evalf(eval(x^1000-1000^x-1, x=999));
evalf(eval(x^1000-1000^x-1, x=1001));

                                           fsolve(x^1000-1000^x-1, x, 999 .. 1001)
                                                        3.666954248*10^2999
                                                       -9.972830761*10^3002

@BennyMopps  Back quotes are used to get the prefix notation of a operator.

Examples:

`$`(a,4);
a$4;
`+`(a,b);
`*`(a,b,c);

@robertocooper 

From help:  "The name  `if`  is an alias for  ifelse . When using this name,  if  must be enclosed in back quotes (left single quotes) because  if  is a Maple reserved word".

See help on   ifelse  for detailes.

@Preben Alsholm The correct answer to the question posed depends on the interpretation (positive or negative). I understood this question in the following sense: choose roots in front of which there is a plus sign. If OP meant any other meaning, then let he just clarify the question. For example, you can put the question like this: find the parameters values at which a given root takes a positive (negative) value.