Finding sum of a series: difference of cubes

Question

I am trying to find sum of the infinite series:

$$1^{3}-2^{3}+3^{3}-4^{3}+5^{3}-6^{3} + \ldots$$

I tried to solve it by subtracting sum of even cubes from odd, but that solves only half of the numbers.

Any input is appreciated.

Thank you all (especially Marvis, Manzonni, Gottfried, Mhenni for long and descriptive answers) for you answer. It's my first visit to math.stackexchange.com. Very impressed with all the help. :)

Now, I would like to know how Marvis got to partial sum formula:

$S_n = \sum_{k=0}^{n}(-1)^{k+1} k^3 = \dfrac{1 + (-1)^n(4n^3 + 6n^2-1)}8 = \begin{cases} \dfrac{2n^3+3n^2}4; & n \text{ is even}\\ \dfrac{1-3n^2-2n^3}4; & n \text{ is odd} \end{cases}$

I tried to replicate it, but can't get to it.

Answer 1

This sum does not converge in the usual sense. By usual sense, we mean that if you look at the sequence of partial sums then we get that $$S_n = \sum_{k=0}^{n}(-1)^{k+1} k^3 = \dfrac{1 + (-1)^n(4n^3 + 6n^2-1)}8 = \begin{cases} \dfrac{2n^3+3n^2}4; & n \text{ is even}\\ \dfrac{1-3n^2-2n^3}4; & n \text{ is odd} \end{cases}$$

Hence, $\displaystyle \lim_{n \rightarrow \infty} S_n$ does not exist since $S_{2k} \to \infty$ while $S_{2k+1} \to - \infty$.

That said, convergence of partial sums is by no means the only way to define convergence. A popular way of defining convergence is to look at the Cesàro sum. In this case, the Cesàro summation technique applied once also diverges. You need to do it thrice to get a finite answer.

There are also other regularization techniques like Borel summation, Ramanujan summation which assign a finite value to sums which do not converge. All these different techniques typically assign the same value to any divergent series.

The $\zeta$ regularization technique (which is actually closely related to Ramanujan summation technique) can also be used here to assign a value to the divergent series. Consider $$g(s) = 1 - \frac1{2^s} + \frac1{3^s} - \frac1{4^s} + \cdots$$ and $$f(s) = 1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots$$ The series $g(s)$ converges for Real$(s) > 0$ and converges absolutely for Real$(s) >1$. The series $f(s)$ converges for Real$(s) > 1$. In the region, Real$(s)>1$, we have that \begin{align} g(s) & = 1 - \frac1{2^s} + \frac1{3^s} - \frac1{4^s} + \cdots\\ & = \left(1 + \frac1{2^s} + \frac1{3^s} + \frac1{4^s} + \cdots \right) - \left(\frac2{2^s} + \frac2{4^s} + \frac2{6^s} + \frac2{8^s} + \cdots \right)\\ & = f(s) - \frac1{2^{s-1}} f(s) = \left(1 - 2^{1-s} \right)f(s). \end{align}

Now analytic continuation of $g(s)$ gives us $\eta(s)$ such that $\eta(s) = (1-2^{1-s}) \zeta(s)$, where $\zeta(s)$ is the analytic continuation of $f(s)$. Hence, plugging in $s=-3$, we get that $$1^3 - 2^3 + 3^3 - 4^3 +\cdots \underset{\text{ac}}{=} \eta(-3) = (1-2^{1-(-3)}) \zeta(-3) = -15\zeta(-3) = -15 \times \dfrac1{120} = -\dfrac18$$

Answer 2

Your series does of course not converge but you may try to 'regularize' it.

Let's replace the power of $3$ by any complex $-z$ to get : $$\eta(z)=-\sum_{k=1}^\infty \frac {(-1)^k}{k^z}$$

This is the Dirichlet eta function $\ \eta(z)=(1-2^{1-z})\zeta(z)\ $ and the series is convergent for any complex $z$ such that $\Re(z)>0$. By analytic continuation we may extend $\zeta$ (and $\eta$ as well) to the whole complex plane (with a simple pole at $z=1$).

Since $z=-3$ in your case you could consider the 'regularized answer' to be $$\ (1-2^{4})\zeta(-3)=-15\,\zeta(-3)$$

$\zeta(-3)$ (as well as other negative integers) may be written using Bernoulli numbers as $\zeta(-3)=-\frac{B_{n+1}}{n+1}=\frac 1{120}$ so that the answer should be $-\frac 18$
(but Marvis +1 got that part a little earlier...)

Answer 3

A related problem. Here is another approach. Applying the operator $(xD)^3=(xD)(xD)(xD)$, where $D=\frac{d}{dx}$, to the equation

$$ \sum_{k=1}^{\infty}(-1)^{k+1}x^k = \frac{x}{(1+x)},$$

gives

$$ \sum_{k=1}^{\infty}(-1)^{k+1}k^3\, x^{k} = \frac{x(1-4x+x^2)}{(1+x)^4}\,. $$

Taking $x=1$, see Abel's theorem, in the above equation, we get the desired result

$$ \sum_{k=1}^{\infty}(-1)^{k+1}k^3 = -\frac{1}{8}\,. $$.

Note that:

$$ \sum_{k=0}^{\infty} (-1)^kx^k=\frac{1}{1+x}. $$

Answer 4

Using the identities $$ \sum_{k=0}^\infty\binom{k}{n}x^k=\frac{x^n}{(1-x)^{n+1}} $$ and $$ k^3=\binom{k}{1}+6\binom{k}{2}+6\binom{k}{3} $$ yields $$ \sum_{k=1}^\infty k^3x^{k-1}=\frac{1}{(1-x)^2}+6\frac{x}{(1-x)^3}+6\frac{x^2}{(1-x)^4} $$ Letting $x\to-1$, gives $$ \begin{align} \sum_{k=0}^\infty(-1)^{k-1}k^3 &=\frac14-6\cdot\frac18+6\cdot\frac1{16}\\ &=-\frac18 \end{align} $$

Answer 5

Another apporach, using the binomial-theorem in a infinitely repeated application.
It has also the advantage, that it does not need the external knowledge of the value of $\zeta(-3)$ and also can be generalized to any positive power in the exponents of the series.

Consider a type of Vandermonde-vector $V(x)=[1,x,x^2,x^3] \quad $ . Then consider the upper triangular pascalmatrix of appropriate size $$ P=\begin{bmatrix} 1&1&1&1\\ .&1&2&3 \\ .&.&1&3 \\ .&.&.&1 \end{bmatrix} $$ Then the matrixproduct holds $$V(x) \cdot P = V(x+1) $$ or, for example $$\begin{matrix} & \cdot & \begin{bmatrix} 1&1&1&1\\ .&1&2&3 \\ .&.&1&3 \\ .&.&.&1 \end{bmatrix} \\ \begin{bmatrix}1 &3 &3^2&3^3 \end{bmatrix} &=& \begin{bmatrix}1&4&4^2&4^3 \end{bmatrix} \end{matrix}$$

Because auf the properties of composition of linear operators we can write, for instance to sum four terms: $$ V(x) \cdot (I - P + P^2 - P^3) = V(x) - V(x+1) + V(x+2)-V(x+3) $$

If it is now possible to express the infinite series of powers of P by the closed form for geometric series for matrices :
$$ I - P + P^2 - P^3 + ... - ... = (I + P)^{-1} = H $$ then we can make the ansatz and deduce formally: $$ \begin{array} {rcl} V(1) \cdot H &= &V(1) - V(2) + V(3) - ... + ... \\ & = & [1,1,1,1] - [1,2,4,8] + [1,3,9,27] - [1,4,16,64] +... - ... \\ & =& \begin{bmatrix}\eta(0), &\eta(-1), &\eta(-2), &\eta(-3) \end{bmatrix} \\ \end{array}$$ where $\eta(m)$ is the Dirichlet's eta-function.

The usability of the closed form for the geometric series also for the geometric matrix-series should be proven first - I do not attempt it here, but such proofs exist: you might follow for instance the chapter Neumann series in wikipedia, but there are a couple of other online resources too.

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Here we simply use, that this ansatz works and we can now compute H to dimension 4x4
$$ \small H= \begin{bmatrix} 2&1&1&1\\ .&2&2&3 \\ .&.&2&3 \\ .&.&.&2 \end{bmatrix} ^{-1} = \small \begin{bmatrix} 1/2 & -1/4 & 0 & 1/8 \\ . & 1/2 & -1/2 & 0 \\ . & . & 1/2 & -3/4 \\ . & . & . & 1/2 \end{bmatrix}$$

and then compute $$ \small \begin{array}{} [1,1,1,1] \cdot H & =&\begin{bmatrix}1/2, & 1/4, & 0, & -1/8 \end{bmatrix} \end{array} $$

In the last column of the result-vector is the value $-\frac18 = \eta(-3)$ which you're asking for.

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[update]: It might be interesting, that this scheme can even made be more convenient for the actual computation in that we do not need the explicite inverse H.
First I show the - a bit lengthy- formal derivation and the simple computation scheme follows below.

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Let's write $G=(I+P)$ because we do not need the inverse H.
We can rearrange the matrix equation, writing $$\small \begin{array}{rcl} Z&=&\begin{bmatrix}\eta(0), &\eta(-1), &\eta(-2), &\eta(-3) \end{bmatrix} \\ Z \cdot G& = &V(1) \\ Z \cdot (G - 2I)& =& V(1)-2Z \\ 1 \cdot V(1) - Z \cdot (G - 2I)& =& 2Z \\ \text{ defining } \\ Z^*&=& \begin{bmatrix}1, \eta(0), &\eta(-1), &\eta(-2), &\eta(-3) \end{bmatrix} \\ G^*&=& \operatorname{concat}(V(1),-(G-2I)) \\ \text{ then it holds } \\ Z^* \cdot G^* &=& 2Z^* \end{array}$$ But then $G^*$ is nilpotent, and powers of it degenerate to a form, where the first row remains nonzero only. If we choose a certain dimension d for G then on the lhs the entries in $Z^*$ become insignificant except the leading 1 and we can write: $$ V(0) \cdot {G^*}^d = 2^d Z^* $$

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That derivation shows, that we can use a repeated very simple doable matrix-multiplication to compute iteratively the sequence of $\eta(-k)$-values. We use the matrix $$ \small G^* = \begin{bmatrix} 2 & 1 & 1 & 1 & 1 & 1 & 1 & \cdots \\ . & 0 & -1 & -1 & -1 & -1 & -1 & \cdots \\ . & . & 0 & -2 & -3 & -4 & -5& \cdots \\ . & . & . & 0 & -3 & -6 & -10 & \cdots \\ . & . & . & . & 0 & -4 & -10 & \cdots \\ . & . & . & . & . & 0 & - 5 & \cdots \\ . & . & . & . & . & . & 0 & \cdots \\ \vdots& \vdots& \vdots&\vdots&\vdots&\vdots&\vdots& \ddots\\ \end{bmatrix}$$ only to the required dimension(!) and compute iteratively (which can be done by paper&pen only) up to the d'th multiplication $$ ([1,0,0,0...]\cdot G^*)\cdot G^* )\cdot \ldots )\cdot G^* = 2^d [1,\eta(0),\eta(-1),...,\eta(1-d), errors] $$

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Since you mention, you'd like any input, you might also be interested in a more worked discussion of this here where I initially worked this out and extended the same principle also to non-alternating sums.