Evaluating $ \sum_{a \in \mathbb{N}}\sum_{b \in \mathbb{N}}\sum_{c \in \mathbb{N}} \frac{(-1)^{a+b+c}}{1+a+b+c}. $

Question

A friend of mine gave me the task to calculate the value of the following triple sum: $$ \sum_{a \in \mathbb{N}}\sum_{b \in \mathbb{N}}\sum_{c \in \mathbb{N}} \frac{(-1)^{a+b+c}}{1+a+b+c}. $$ I am not even sure whether or not it converges. I have tried to interchange the order of summation and just calculate the value of $$ \sum_{n \in \mathbb{N}} \frac{\phi(n) \cdot (-1)^n}{1 + n} $$ where $\phi(n)$ is the number of ways $n$ can be written as the sum of positive integers, but this sum clearly diverges.

So I wonder

• Does this triple sum converge?
• If so, what value does it converge to?

Answer 1

$$\begin{split} S_3 &:=\sum_{a=0}^\infty\sum_{b=0}^\infty\sum_{c=0}^\infty \frac{(-1)^{a+b+c}}{1+a+b+c} \\ &=\sum_{a=0}^\infty\sum_{b=0}^\infty\sum_{c=0}^\infty(-1)^{a+b+c}\int_0^1x^{a+b+c}\,dx\\ &=\int_0^1\sum_{a=0}^\infty(-x)^a\cdot\sum_{b=0}^\infty(-x)^b\cdot\sum_{c=0}^\infty(-x)^c\, dx\\ &=\int_0^1 \frac{dx}{(1+x)^3} = \bbox[5px,border:2px solid red]{ \frac{3}{8}}\,. \end{split}$$ Can you fill in the details? :)

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More generally, for integer $k\geqslant 2$, $$\begin{split} S_k&:=\sum_{a_1=0}^\infty\cdots\sum_{a_k=0}^\infty\frac{(-1)^{\sum_{j=1}^ka_j}}{1+\sum_{j=1}^ka_j}\\ &=\int_0^1\prod_{j=1}^k\sum_{a_j=0}^\infty (-x)^{a_j}dx\\ &=\int_0^1\frac{dx}{(1+x)^k}= \bbox[5px,border:2px solid red]{\frac{1-2^{1-k}}{k-1}}\,. \end{split}$$ Continuing the expression at $k=1$, we retrieve the well-known result $$S_1:=\sum_{n=0}^\infty \frac{(-1)^n}{1+n}=\log(2)\,. $$

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Bonus I. Just for fun: $$\begin{split} \sum_{k=1}^n S_k &= \log(2) - \sum_{k=2}^n \frac{2^{1-k}}{k-1} + \sum_{k=2}^n \frac{1}{k-1} \\ &=\log(2) - \sum_{k=1}^n \frac{1}{k\cdot 2^k} + \sum_{k=1}^n \frac 1 k\,. \end{split}$$ However, $$-\log(1-x) = \sum_{k=1}^\infty \frac{x^k}{k}\,,\qquad |x|<1\,, \tag{$\ast$}$$ so that $$\sum_{k=1}^n \frac{1}{k\cdot 2^k} \xrightarrow{n\to\infty} -\log\left(1-\frac 1 2\right) = \log(2)\,, $$ that is, $$\log(2) - \sum_{k=1}^n \frac{1}{k\cdot 2^k} = \mathrm{o}(1)\qquad \text{as }n\to\infty\,. $$ As a consequence $$\begin{split} \sum_{k=1}^n S_k &= H_n + \mathrm{o}(1) \\ &=\log(n) + \gamma + \mathrm{o}(1)\qquad \text{as }n\to\infty\,, \end{split}$$ where $H_n$ is the $n$-th harmonic number, and $\gamma$ is the Euler–Mascheroni constant. These asymptotics can be sharpened by writing down Lagrange's remainder theorem for $(\ast)$ and by refining the expansion for $H_n$, for example via the Euler–Maclaurin formula or its equivalents (see answers here).

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Bonus II. (Sorry, I'm bored on a long train journey.) Same thing but with doubled integers in the denominator: $$\begin{split} R_k &:= \sum_{a_1=0}^\infty \cdots \sum_{a_k=0}^\infty \frac{(-1)^{\sum_{j=1}^k a_j}}{1+\color{green}{2}\sum_{j=1}^k a_j} \\ &= \int_0^1 \prod_{j=1}^k \sum_{a_j=1}^\infty (-1)^{a_j} x^{2a_j}\,dx\\ &= \int_0^1 \frac 1{(x^2+1)^k}\, dx = \int_0^{\pi/4} \cos^{2k-2}(x)\,dx \end{split}$$ We immediately deduce $R_1 = \frac \pi 4$, in accordance with $$\sum_{a=0}^\infty \frac{(-1)^a}{1+2a} = \arctan(1) = \frac{\pi}4\,.$$ For integer $k\geqslant 2$, integrating by parts yields the recurrence relation $$\begin{split} R_k &=\int_0^{\pi/4} \cos^{2k-2}(x)\,dx = \left.\frac{\sin(x)\cdot\cos^{2k-3}(x)}{2k-2}\right|_0^{\pi/4}+ \frac{2k-3}{2k-2}\int_0^{\pi/4} \cos^{2k-4}(x)\,dx\\ &= \frac{1}{2^k\cdot(k-1)} + \frac{2k-3}{2k-2}\,R_{k-1}\,. \end{split}$$ Solving for $R_k$ is complicated (maybe Mathematica can give you a closed-form expression), but at the very least we can compute $$R_3 = \frac{1}{16} + \frac{3}{4}\,R_2 =\cdots= \frac{8+3\pi}{32}\,.$$ Not sure whether you can do this with numbers other than $2$ in that denominator.

Answer 2

I believe the sum does converge, as does the analogous $k$-fold sum for any $k\ge1$. Let me try to make a convincing argument for $k=2$. We can write $$ \sum_{a=0}^\infty \sum_{b=0}^\infty \frac{(-1)^{a+b}}{1+a+b} = \sum_{a=0}^\infty (-1)^a T(a), \quad\text{where}\quad T(a) = \sum_{b=0}^\infty \frac{(-1)^b}{1+a+b}. $$ Note that $T(a)$ converges by the alternating series test. We'd like to use the alternating series test on the remaining sum over $a$; but to do so, we need to show that $T(a)$ decreases to $0$. The fact that $\lim_{a\to\infty} T(a)\to0$ isn't hard, because $T(a)$ is bounded between its initial partial sum $0$ and its $0$th partial sum $\frac1{1+a}$. It's “decreasing” that we need to work for.

We want to say that $$ T'(a) = \sum_{b=0}^\infty \frac d{da} \frac{(-1)^b}{1+a+b} = - \sum_{b=0}^\infty \frac{(-1)^b}{(1+a+b)^2} $$ because this would show that $T'(a)<0$ by the same alternating series argument (it's bounded between $0$ and the $0$th partial sum $-\frac1{(1+a)^2}$). However, term-by-term differentiation of an infinite series needs to be justified: the series of derivatives must converge (locally) uniformly. Fortunately that is the case here since the series of derivatives converges absolutely and uniformly bounded by $\sum_{b=0}^\infty \frac1{(1+b)^2}$.

My instinct is that this argument can be iterated to the original triple sum (and beyond), but I haven't tried to write down the details.

Answer 3

THIS IS NOT A COMPLETE ANSWER!

I provide an approach that does work to provide the answer, although I do not have the knowledge of tools required to justify convergence issues. Let us concern ourselves with finding

$$S(x)=\sum_{a=0}^{\infty}\sum_{b=0}^{\infty}\sum_{c=0}^{\infty} \frac{x^{a+b+c}}{1+a+b+c}$$

Then the ordered pairs of non-negative integers $(a,b,c)$ such that $a+b+c=n$ is $\binom{n+2}{2}$ by stars and bars and so your sum can be written as

$$S(x)= \sum_{r=0}^{\infty} \binom{r+2}{2} \frac{x^r}{1+r} = \sum_{\color{blue}{r=1}}^{\infty} \binom{r+1}{2} \frac{x^{r-1}}{r} =\sum_{r=1}^{\infty} \binom{r+1}{r} \frac{x^r}{2x} = \frac1{2x}\left(\sum_{r=0}^{\infty} \binom{2+r-1}{r} x^r\right) - \frac1{2x}$$

This is the binomial series for $(1-x)^{-2}$ which converges for $|x|<1$. Thus $2xS(x)=\frac1{(1-x)^2}-1 = \frac{2x-x^2}{(x-1)^2}$. This gives $$S(x)=\frac{2-x}{2(1-x)^2}$$

And although this formula is only valid for $x\in (-1,1)$, putting $x=-1$ yields $S(-1)=\frac38$ which happens to be correct as per the calculus-based answer. I am not sure why this works but a little digging on the internet led me to Abel’s theorem. Perhaps some one with enough knowledge could use it to justify the steps. Or perhaps this idea is completely incorrect. I’d appreciate if someone could comment on the same.

Answer 4

Since the question of the value of the sum has already been addressed, let me tackle the question about the convergence.

Lemma. For any $a \in \mathbb{C} \setminus\{0,-1,-2,\ldots\}$,

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{k+a} = \sum_{n=0}^{\infty} \frac{n!}{2^{n+1}a(a+1)\cdots(a+n)}.$$

Also, for any given positive integer $N$, there exist constants $c_0, \ldots, c_{N-1}$ such

$$\sum_{k=0}^{\infty} \frac{(-1)^k}{k+a} = \sum_{n=0}^{N-1} \frac{c_n}{a+n} + \mathcal{O}(a^{-N-1}) $$

as $\operatorname{Re}(a) \to \infty$.

I think this can be directly obtained by the Euler's series acceleration, although a direct proof is available (see below). Repeatedly applying this lemma then proves that OP's series converges.

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Proof of Lemma. Assume first that $\operatorname{Re}(a) > 0$. For a given positive integer $K$,

$$ \begin{align*} \sum_{k=0}^{K-1} \frac{(-1)^k}{k+a} &= \sum_{k=0}^{K-1} (-1)^k \int_{0}^{1} x^{k+a-1} \, \mathrm{d}x \\ &= \int_{0}^{1} \sum_{k=0}^{K-1} (-1)^k x^{k+a-1} \, \mathrm{d}x \\ &= \int_{0}^{1} \frac{1 - (-x)^K}{1 + x} x^{a-1} \, \mathrm{d}x \\ &\to \int_{0}^{1} \frac{x^{a-1}}{1 + x} \, \mathrm{d}x \end{align*} $$

as $K \to \infty$ by the dominated convergence theorem. Now, by performing integration by parts, we get

$$ \begin{align*} \int_{0}^{1} \frac{x^{a-1}}{1 + x} \, \mathrm{d}x &= \left[ \frac{x^a}{a(x+1)} \right]_{0}^{1} + \frac{1}{a} \int_{0}^{1} \frac{x^{a}}{(x+1)^2} \, \mathrm{d}x \\ &= \frac{1}{2a} + \frac{1}{2^2 a(a+1)} + \frac{2!}{a(a+1)} \int_{0}^{1} \frac{x^{a+1}}{(x+1)^3} \, \mathrm{d}x \\ &\qquad \vdots \\ &= \sum_{n=0}^{N-1} \frac{n!}{2^{n+1}a(a+1)\cdots(a+n)} + \frac{N!}{a(a+1)\cdots(a+N-1)} \int_{0}^{1} \frac{x^{a+N-1}}{(x+1)^N} \, \mathrm{d}x \tag{*} \end{align*} $$

Since the factor $\frac{N!}{a(a+1)\cdots(a+N-1)}$ grow at most polynomially fast and $0 \leq \frac{x}{x+1} \leq \frac{1}{2}$ for $0 \leq x \leq 1$, the remainder term converges to $0$ as $N \to \infty$, proving the first assertion of the lemma when $\operatorname{Re}(a) > 0$. The general case then easily follows from the principle of analytic continuation.

The second assertion also easily follows from the formula $\text{(*)}$, first expanding the summation part using the partial fraction decomposition and then noting that the remainder term decays at least as fast as $a^{-N}$ as $\operatorname{Re}(a) \to \infty$ for each fixed $N$.

Answer 5

I think your idea is good. Assuming your sums don't include 0, you can actually use stars and bars to give an expression for $\Phi(n)$:

$\Phi(n) = \binom{n-1}{2}$. If you are summing over the 0s as well (i.e. if $(0,0,n)$ is a valid triple) then that will become $\Phi(n) = \binom{n+2}{2}$.

Now plug that into your expression and you get

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

Obviously this does not converge.

Answer 6

With no proof

If $k$ is the number of $\sum_{0}^\infty$, the result for $k>1$ is given by $$\large\frac {a_k}{2\,b_k}$$ where the $a_k$ and $b_k$ respectively correspond to sequences $A090633$ and $A090634$ in $OEIS$.

The first ones are $$\left\{\frac{1}{2},\frac{3}{8},\frac{7}{24},\frac{15}{64},\frac{31}{160},\frac{21}{128},\frac{127}{896}, \frac{255}{2048},\frac{511}{4608}\right\}$$