Maple readily turns partial sum 1+sum(alpha^2, j = 1 .. l-1) into expression 1+(l-1)*alpha^2. 

I need to replace alpha with a function c(j) such that c(0)=1 and c(j)=alpha for j=1,2,3,...and to compute sum(c(j)^2, j = 0 .. l-1) symbolically to get the above expression. This is the simplest special case only and in general c(j) is 1 for j=0 and some expression involving j and alpha, beta, etc for j::posint.

I tried several thing, e.g., 

restart; with(Statistics)

#Define:

A := RandomVariable(Geometric(p))+1; F := Mean(A);M := abs(A-F)/A;

#and ask Maple to find a pdf of the r.v. M:

f := `assuming`([PDF(M, x)], [x > 0])

#Maple 15 gives an answer, but what does it mean by the output for this input

subs(p = 1/3, x = 3, f)

#or how do I plot this pdf of M? Does Maple give a correct answer?

# Why does the error message appear when I do subs(p = 1/3, x < 3, f...

I wonder how do I show with Maple that for

p:=(x,a,b)->a*(a+x*b)^(x-1)*(exp(-(a+x*b)))/(x!);

the series

sum(p(x,a,b),x=0..infinity) assuming a>0,b<1,b>-1

converges to 1. I also tried

sum(a*(a+x*b)^(x-1)*(exp(-(a+x*b)))/(x!), x=m..infinity) assuming a>0,b<1,b>-1,a+m*b<=0

but all to no avail. For b=0, Maple shows the series converges and we have the Poisson distribution. For b in (-1,1), the (discrete) density...

Why does this happen to Maple 15?

    `assuming`([sum(k*p*Beta(k, p+1), k = 1 .. infinity)], [p > 1]); eval(%, p = 2)

Is it possible to show with Maple that for any real p>1 the series converges to p/(p-1), e.g.,

    `assuming`([sum(k*p*Beta(k, p+1), k = 1 .. 1000)], [p > 1]): subs(p = 2, %): evalf(%)

How do I show this symbolically? Thanks.

Maple evaluates

> sum(1/z^2, z = 1 .. infinity)

to (1/6)*Pi^2. What do I do when sumation is over all even positive integers? Is there any  closed-form symbolic formula for this case?

Above is the minimal example. What I need is to compute 

> sum(2*binomial(m-1, k-1)*binomial(n-1, k-1)/binomial(m+n, m), k = 1 .. infinity)

where k belongs to a set of only even positive integers, not any posint. The second expression evaluates to 2*m*n/((m+n-1...