Formalizing a proof for $ \sum_{n=0}^\infty \sum_{k=n}^{\infty} a_k = \sum_{n=0}^{\infty} (n+1)a_n$

Question

Let $(a_n)_n$ be a sequence of non-negative real numbers.

I fail to see a formal proof of the following (conjectured) equality: $$ \sum_{n=0}^\infty \sum_{k=n}^{\infty} a_k = \sum_{n=0}^{\infty} (n+1)a_n. $$

Roughly speaking the claim follows from expanding the LHS: $$ \sum_{n=0}^\infty \sum_{k=n}^{\infty} a_k = \sum_{k=0}^{\infty} a_k + \sum_{k=1}^{\infty} a_k + \ldots + \sum_{k=i}^{\infty} a_k + \ldots $$ Now it is easy to see that each term $a_k$ appears in exactly $(k+1)$ terms in RHS (e.g. writing column-wise the different sums), so it is intuitively clear that it has to be $\sum_{n=0}^{\infty} (n+1)a_n$.

However I am not satisfied of this informal hand-waving proof. Any hints/helps to make it formal? Thanks.

Answer 1

We can use the indicator function

$$\mathbf{1}_{\{k\geqslant n\}} = \begin{cases}1,&k \geqslant n \\0, &k < n\end{cases},$$

and nonnegativity of $a_k$ to obtain

$$\sum_{n = 0}^\infty\sum_{k=n}^\infty a_k = \sum_{n = 0}^\infty\sum_{k=0}^\infty \mathbf{1}_{\{k\geqslant n\}}a_k \,\underbrace{= \sum_{k = 0}^\infty\sum_{n=0}^\infty \mathbf{1}_{\{k\geqslant n\}}a_k }_{\text{applying Tonelli's theorem}} = \sum_{k = 0}^\infty a_k\sum_{n=0}^\infty \mathbf{1}_{\{k\geqslant n\}} \\ = \sum_{k = 0}^\infty a_k\sum_{n=0}^k (1) = \sum_{k=0}^\infty (k+1)a_k = \sum_{n=0}^\infty (n+1)a_n$$ Clearly, the last step is just replacing letter "k" with letter "n".

Answer 2

By double counting, since the sum by columns must be equal to the sum by rows, we obtain that

$$\sum_{n=0}^\infty \sum_{k=n}^{\infty} a_k = \sum_{k=0}^{\infty}\sum_{n=0}^{k} a_k= \sum_{k=0}^{\infty}a_k\sum_{n=0}^{k}1 =\sum_{k=0}^{\infty}(k+1)a_k$$

Refer also to the related

• How to change the order of summation?

Answer 3

I think it's actually less trouble to prove Tonelli's theorem (which I imagine in this context means something like Theorem 8.43 of Apostol, Mathematical Analysis (2nd ed. 1974), although the more general* proposition (5.3.6) of Dieudonné, Foundations of Modern Analysis (1969) could also be used) than it is to prove this result without using Tonelli's theorem! However, nothing daunted $\ldots$

^{*Minor correction (not needed for this argument): it isn't strictly accurate to describe Dieudonne's (5.3.6) as more general than Apostol's Theorem 8.43. I listed some other references in a comment on this related question.}

Define the triangular numbers: $$ t_k = \frac{k(k+1)}{2} \quad (k \geqslant 0), $$ and the infinite sequence $(b_r)_{r\geqslant0}$, where: $$ b_r = a_k \quad (k \geqslant 0,\ t_k \leqslant r < t_{k+1}), $$ thus: $$ \begin{array}{c|ccccccccccc} r & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \cdots \\ b_r & a_0 & a_1 & a_1 & a_2 & a_2 & a_2 & a_3 & a_3 & a_3 & a_3 & \cdots \end{array} $$ For $N \geqslant 0$, define the $(t_{N+1})^\text{th}$ partial sum of the series $\sum_{r=0}^\infty b_r$, thus: $$ s_N = \sum_{r=0}^{t_{N+1}-1}b_r = \sum_{k=0}^N(k+1)a_k \quad (N \geqslant 0). $$ Consider this infinite triangular array: $$ \begin{array}{c|ccccc} \vdots & \vdots & \vdots & \vdots & \vdots & \\ \text{row } 3 & & & & b_9 & \cdots \\ \text{row } 2 & & & b_5 & b_8 & \cdots \\ \text{row } 1 & & b_2 & b_4 & b_7 & \cdots \\ \text{row } 0 & b_0 & b_1 & b_3 & b_6 & \cdots \\ \hline \text{sums} & a_0 & 2a_1 & 3a_2 & 4a_3 & \cdots \end{array} $$ The $n^\text{th}$ row of the array is the infinite sequence $\big(c_k^{(n)}\big)_{k\geqslant{n}}$, where: $$ c_k^{(n)} = b_{t_k + n} = a_k \quad (k \geqslant n \geqslant 0). $$ In terms of the $c_k^{(n)}$, the selected partial sums of the series $(b_r)_{r\geqslant0}$ are given by: $$ s_N = \sum_{n=0}^N\sum_{k=n}^Nc_k^{(n)} \quad (N \geqslant 0). $$ For $N \geqslant n \geqslant 0$, the $(N-n)^\text{th}$ remainder of the infinite series $\sum_{k=n}^{\infty}c_k^{(n)}$ is the sum of a subsequence of $(b_r)$ whose indices satisfy $r\geqslant{t_{N+1}}$. For $n > N \geqslant 0$, the entire infinite series $\sum_{k=n}^{\infty}c_k^{(n)}$ is the sum of such a subsequence of $(b_r)$. Also, $$ t_k + n \ne t_{k'} + n' \quad (k \geqslant n \geqslant 0, \ k' \geqslant n' \geqslant 0, \ n \ne n'). $$ This means that all the series just mentioned are the sums of pairwise disjoint subsequences of the sequence $(b_r)$ with indices $r\geqslant{t_{N+1}}$.

Now we begin to consider questions of convergence. It is required to prove: $$ \sum_{n=0}^\infty \sum_{k=n}^{\infty} a_k = \sum_{n=0}^{\infty} (n+1)a_n, $$ where $a_n \geqslant 0$ for all $n$. This is trivial if $a_n = 0$ for all $n$, so we assume from now on that $a_n > 0$ for some $n$, whence the sums on either side of the identity, if well-defined, are also strictly positive. If the iterated infinite sum on the left hand side is well-defined and equal to $S > 0$, then $S$ is an upper bound for all the finite sums: $$ \sum_{n=0}^N\sum_{k=n}^N a_k = \sum_{n=0}^N(n+1)a_n \quad (N \geqslant 0), $$ whence the infinite sum on the right hand side of the conjectured identity is also well-defined.

It is therefore enough to prove that if the sum on the right hand side of the identity is well-defined and equal to $S > 0$, then the iterated sum on the left hand side of the identity is also well-defined and equal to $S$.

Given $\epsilon > 0$, choose $N$ so that: $$ \sum_{r=t_{N+1}}^\infty b_r = \sum_{n=N+1}^\infty(n+1)a_n \leqslant \epsilon. $$

Lemma Let $m$ be an integer, and let $(b_r)_{k \geqslant m}$ be a sequence of non-negative numbers such that $\sum_{r=m}^\infty b_r$ converges. For any infinite set $$ L = \{l(0), l(1), l(2), \ldots\} \text{ where } m \leqslant l(0) < l(1) < l(2) < \cdots, $$ define $$ \sigma(L) = \sum_{i=0}^\infty b_{l(i)}. $$ Then for any infinite sequence of pairwise disjoint infinite subsets $L_0, L_1, L_2 \ldots$ of $\{r \in \mathbb{Z} : r \geqslant m\}$ we have $$ \sum_{n=0}^\infty \sigma(L_n) \leqslant \sum_{r=m}^\infty b_r. $$ Proof In the contrary case, there are $\eta > 0$ and an integer $M \geqslant 1$ such that $$ \sum_{n=0}^{M-1} \sigma(L_n) = \sum_{r=m}^\infty b_r + 2\eta. $$ From $L_n$ ($n = 0, 1, \ldots, M-1$) take enough terms of the series for $\sigma(L_n)$ to give a partial sum greater than $\sigma(L_n) - \eta/M$. By the disjointness condition, all these terms together have a sum greater than $$ \sum_{n=0}^{M-1} \sigma(L_n) - \eta > \sum_{r=m}^\infty b_r, $$ which is a contradiction.

The lemma shows that the $(N-n)^\text{th}$ remainder of the infinite series $\sum_{k=n}^{\infty}c_k^{(n)}$ for $n = 0, 1, \ldots, N$, together with the entirety of the series $\sum_{k=n}^{\infty}c_k^{(n)}$ for $n = N+1, N+2, \ldots$, form an infinite series whose sum is at most $\epsilon$. Denote this by $\sum_{n=0}^\infty d_n$. Then: $$ \sum_{n=0}^N\sum_{k=n}^\infty c_k^{(n)} = \sum_{n=0}^N\left(\sum_{k=n}^N c_k^{(n)} + d_n\right) = \sum_{n=0}^N\sum_{k=n}^N c_k^{(n)} + \sum_{n=0}^N d_n, $$ therefore: $$ \sum_{n=0}^\infty\sum_{k=n}^\infty a_k = \sum_{n=0}^\infty\sum_{k=n}^\infty c_k^{(n)} = \sum_{n=0}^N\sum_{k=n}^N c_k^{(n)} + \sum_{n=0}^\infty d_n = s_N+ \sum_{n=0}^\infty d_n. $$ This shows that the iterated sum on the left hand side of the conjectured identity is well-defined, and its value lies between $s_N$ and $s_N + \epsilon$. The two sides of the identity therefore differ by most $\epsilon$. Since $\epsilon$ was arbitrary, this proves that they are equal.