Question

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!)

Answer 1

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)`

Answer 2

`(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`