Compute the following? #pi^4+4pi^3i-10pi^2-20pi*i+35+(56i)/pi-84/(pi^2)-(120i)/(pi^3)+(165)/(pi^4)...#

I made this problem, and so with the solution, could you please tell me how difficult the problem was? (And please post as many different solutions possible!)

2 Answers
Apr 6, 2018

The sum is of the form

#sum_(n=0)^oo pi^4(i /pi)^n t_n#

where the #t_n# form the sequence

1,4, 10,20, 35,56,...

Short approach

Each term in the sequence is #1/6# th of the corresponding terms
of the sequence

# 6,24,60,120, 210, 336,...#

which is the sequence

# 1times 2 times 3, 2times 3 times 4, 3times 4 times 5, 5times 6 times 7, 6 times 7 times 8,...#

Thus #t_n = 1/6 (n+1)(n+2)(n+3)#

So, the sum is

#S = pi^4/6 sum_{n=0}^oo (n+1)(n+2)(n+3)rho^n#

where #rho = i/pi#. Now

#S = pi^4/(6rho^3) sum_{n=0}^oo (n+1)(n+2)(n+3)rho^(n+3)#
#qquad = pi^4/(6rho^3) sum_{n=0}^oo rho^3 d^3/(d rho^3)rho^(n+3) = pi^4/(6) d^3/(d rho^3)[sum_{n=0}^oo rho^(n+3)]#
#qquad = pi^4/(6) d^3/(d rho^3)[rho^3/(1-rho)] = pi^4/(6) d^3/(d rho^3)[1/(1-rho)-1-rho-rho^2]#
#qquad = pi^4/6 (3!)/(1-rho)^4 = color(red)(pi^4/(1-rho)^4)#
#= (pi/(1-i/pi))^4=pi^8/(pi-i)^4#

Long approach

It may not always be possible to determine the form of the sequence by inspection. To tackle the problem under this situation, let us form a table of successive differences

#((1,4,10,20,35,56,...),(3,6,10,15,21,...,...), (3,4,5,6,...,...,....),(1,1,1,...,....,....,...))#

The table shows clearly that the fourth (and all further) successive difference of the terms vanish.

It is convenient to use so called factorial polynomials, defined by

#n^((r)) equiv n(n-1)(n-2)...(n-r+1), qquad r in NN#
#n^((0)) equiv 1#

These obey the difference relation

#Delta n^((r)) equiv (n+1)^((r)) - n^((r)) =r n^((r-1))#

Note the similarity between this and the corresponding formula for differentiating #n^r#. We can exploit this similarity to develop an expansion similar to Taylor series.

If we write

#t_n = sum_(r=0)^oo c_r n^((r))#

it is easy to see that

#c_r = (Delta^r t_0)/(r!)#

In our case, we can see from the table that

#c_0 =1, c_1=3, c_2=3/2, c_3=1/6#

are the only nonzero coefficients.

Now, let us consider the sum

#S_r equiv sum_(n=0)^oo n^((r)) rho^n#

Note that for #r >= 1# we can change the lower limit of the sum to #r#, the first #r-1# terms being zero. This sum obeys

#rho S_r =sum_(n=0)^oo n^((r)) rho^(n+1) = sum_(n=1)^oo (n-1)^((r)) rho^n#

So, for #r>= 1#

#(1-rho)S_r = sum_(n=1)^oo Delta((n-1)^((r))) rho^n#
#qquad =rho sum_(n=1)^oo r(n-1)^((r-1)) rho^n #
#qquad = rho r sum_(n=0)^oo n^((r-1)) rho^n#
#qquad =rho r S_(r-1)#

and hence we have the recursion relation

#S_r =r rho /(1-rho)S_(r-1)#

Since

#S_0 = 1/(1-rho)#

we can immediately see that

#S_r = r! rho^r/(1-rho)^(r+1)#

(this could have also been obtained by successive differentiation of
#S_0#)
Our sum is

#S = pi^4(S_0+3S_1+3/2S_2+1/6S_3)#

with #rho = i/pi#, which is

#S = pi^4(1/(1-rho) + 3 rho/(1-rho)^2+3/2 2 rho^2/(1-rho)^3+1/6 6 rho^3/(1-rho)^4)#
#qquad = pi^4/(1-rho)^4 ((1-rho)^3+3rho(1-rho)^2+3rho^2(1-rho)+rho^3)#
# qquad = color(red)(pi^4/(1-rho)^4)#

Apr 7, 2018

#(pi/(1-i/pi))^4#

Explanation:

Notice here that this series is in the form #a^4+4a^4r+10a^4r^2+20a^4r^3+...#

Also notice that this is #a+ar+ar^2+ar^3...# raised to the power of 4.

We find that #a^4=pi^4#

#a=pi#

Using this information, we see that #r=i/pi#

The sum of our series is equal to #(pi+i-1/pi-i/pi^2+1/pi^3+...)^4#

The sum of that series is #a/(1-r)#

The sum of our series is therefore #(pi/(1-i/pi))^4#

By the way, this is equivalent to #(pi^8(pi^4-6pi^2+1))/(pi^2+1)^4+(4pi^3(pi^2-1))/(pi^2+1)^4*i#