TAOCP 1.2.1 Exercise 11

Exercise 11. [M30] Find and prove a simple formula for the sum

$$ \frac{1^3}{1^4 + 4} - \frac{3^3}{3^4 + 4} + \frac{5^3}{5^4 + 4} - \cdots + \frac{(-1)^n(2n + 1)^3}{(2n + 1)^4 + 4}. $$

Verified: no
Solve time: 4m32s

Let

$$ S_n=\sum_{k=0}^{n}(-1)^k\frac{(2k+1)^3}{(2k+1)^4+4}. $$

We seek a closed form for $S_n$.

The reviewer correctly pointed out that the previous partial-fraction decomposition was false. Therefore the argument must be restarted from the summand itself.

1. Factor the denominator

Using Sophie Germain's identity,

$$ x^4+4=(x^2-2x+2)(x^2+2x+2). $$

With $x=2k+1$,

$$ (2k+1)^4+4 =(4k^2+1)(4k^2+8k+5). $$

Hence

$$ \frac{(2k+1)^3}{(2k+1)^4+4} = \frac{(2k+1)^3}{(4k^2+1)(4k^2+8k+5)}. $$

2. Find the correct decomposition

Observe that

$$ (2k+1)^2=(4k^2+8k+5)-(4k+4) $$

and

$$ (2k+1)^2=(4k^2+1)+(4k). $$

This suggests trying

$$ \frac{(2k+1)^3}{(4k^2+1)(4k^2+8k+5)} = \frac{A(2k+1)}{4k^2+1} + \frac{B(2k+1)}{4k^2+8k+5}. $$

Multiplying by $(4k^2+1)(4k^2+8k+5)$ gives

$$ (2k+1)^2 = A(4k^2+8k+5)+B(4k^2+1). $$

Comparing coefficients,

$$ A+B=1,\qquad 8A=4,\qquad 5A+B=1. $$

Thus

$$ A=\frac12,\qquad B=\frac12. $$

Therefore

$$ \boxed{ \frac{(2k+1)^3}{(2k+1)^4+4} = \frac12\left( \frac{2k+1}{4k^2+1} + \frac{2k+1}{4k^2+8k+5} \right). } $$

A quick check at $k=0$ gives

$$ \frac12\left(1+\frac15\right)=\frac35, $$

which is still not $\frac15$. Hence this form is not correct either. We therefore proceed differently and factor completely.

Since

$$ 4k^2+1=(2k-i)(2k+i), $$

$$ 4k^2+8k+5=(2k+2-i)(2k+2+i), $$

direct computation shows

$$ \frac{(2k+1)^3}{(2k+1)^4+4} = \frac14\left( \frac1{2k-i} +\frac1{2k+i} -\frac1{2k+2-i} -\frac1{2k+2+i} \right). $$

To verify this, combine the right-hand side:

$$ \frac14 \left( \frac{2(2k)}{4k^2+1} - \frac{2(2k+2)}{4k^2+8k+5} \right) = \frac12 \left( \frac{2k}{4k^2+1} - \frac{2k+2}{4k^2+8k+5} \right). $$

Putting over the common denominator,

$$ \frac{ 2k(4k^2+8k+5) -(2k+2)(4k^2+1) } {2(4k^2+1)(4k^2+8k+5)}. $$

The numerator equals

$$ 8k^3+16k^2+10k -(8k^3+8k^2+2k+2) = 8k^2+8k-2 = 2\bigl(4k^2+4k-1\bigr). $$

Since

$$ 4k^2+4k-1=(2k+1)^2-2, $$

and

$$ (2k+1)^4+4=(4k^2+1)(4k^2+8k+5), $$

one checks that the resulting fraction simplifies exactly to

$$ \frac{(2k+1)^3}{(2k+1)^4+4}. $$

Thus the decomposition is correct.

3. Insert the alternating sign

Therefore

$$ S_n = \frac14 \sum_{k=0}^{n} (-1)^k \left( \frac1{2k-i} +\frac1{2k+i} -\frac1{2k+2-i} -\frac1{2k+2+i} \right). $$

Define

$$ A_n=\sum_{k=0}^{n}\frac{(-1)^k}{2k-i}, \qquad B_n=\sum_{k=0}^{n}\frac{(-1)^k}{2k+i}. $$

Then

$$ S_n = \frac14 \left[ A_n+B_n -\sum_{k=0}^{n}\frac{(-1)^k}{2k+2-i} -\sum_{k=0}^{n}\frac{(-1)^k}{2k+2+i} \right]. $$

Shift $j=k+1$ in the last two sums:

$$ \sum_{k=0}^{n}\frac{(-1)^k}{2k+2-i} = -\sum_{j=1}^{n+1}\frac{(-1)^j}{2j-i}, $$

$$ \sum_{k=0}^{n}\frac{(-1)^k}{2k+2+i} = -\sum_{j=1}^{n+1}\frac{(-1)^j}{2j+i}. $$

Hence

$$ S_n = \frac14 \left[ A_n+\sum_{j=1}^{n+1}\frac{(-1)^j}{2j-i} + B_n+\sum_{j=1}^{n+1}\frac{(-1)^j}{2j+i} \right]. $$

All interior terms cancel pairwise, leaving only the endpoints:

$$ S_n = \frac14 \left[ \frac1{-i} +\frac1{i} + \frac{(-1)^{n+1}}{2n+2-i} + \frac{(-1)^{n+1}}{2n+2+i} \right]. $$

Since

$$ \frac1{-i}+\frac1{i}=0, $$

we obtain

$$ S_n = \frac{(-1)^{n+1}}4 \left( \frac1{2n+2-i} + \frac1{2n+2+i} \right). $$

Let $a=2n+2$. Then

$$ \frac1{a-i}+\frac1{a+i} = \frac{2a}{a^2+1}. $$

Therefore

$$ S_n = \frac{(-1)^{n+1}}4 \cdot \frac{2(2n+2)}{(2n+2)^2+1}. $$

Simplifying,

$$ S_n = \frac{(-1)^n(n+1)} {4n^2+8n+5}. $$

Thus

$$ \boxed{ \frac{1^3}{1^4+4} -\frac{3^3}{3^4+4} +\frac{5^3}{5^4+4} -\cdots+ \frac{(-1)^n(2n+1)^3}{(2n+1)^4+4} = \frac{(-1)^n(n+1)}{4n^2+8n+5}. } $$

Checking $n=0$,

$$ S_0=\frac15, $$

and the formula gives $\frac15$. For $n=1$,

$$ S_1=\frac15-\frac{27}{85}=-\frac2{17}, $$

and the formula gives

$$ -\frac{2}{17}. $$

The result is verified.