Questions tagged [summation-by-parts]

Summation by parts for discrete variables is the equivalent of integration by parts for continuous variables.

Summation by parts transforms the summation of products of sequences into other summations, often simplifying the computation or (especially) estimation of certain types of sums. The summation by parts formula is sometimes called Abel's lemma or Abel transformation.

Suppose $ \lbrace f_{k} \rbrace $ and $ \lbrace g_{k} \rbrace $ are two sequences. Then,

$$ \sum _ {k=m} ^ {n} f _ {k} (g _ {k+1} - g _ {k} )= \left( f _ {n} g _ {n+1} - f _ {m} g _ {m} \right) - \sum _ {k=m+1} ^ {n} g _ {k} (f _ {k} - f _ {k-1} ) \text . $$

Using the forward difference operator $ \Delta $, it can be stated more succinctly as

$$ \sum _ {k=m} ^ {n} f _ {k} \Delta g _ {k} = \left( f _ {n} g _ {n+1} - f _ {m} g _ {m} \right) - \sum _ {k=m} ^ {n-1} g _ {k+1} \Delta f _ {k} \text , $$

Summation by parts is an analogue to integration by parts:

$$ \int f \ dg = f g - \int g \ df \text , $$

or to Abel's summation formula:

$$ \sum _ {k=m+1} ^ {n} f(k) (g _ {k} - g _ {k-1} ) = \left( f(n) g _ {n} - f(m) g _ {m} \right) - \int _ {m} ^ {n} g _ { \lfloor t \rfloor } f'(t) dt \text . $$

An alternative statement is

$$ f _ {n} g _ {n} - f _ {m} g _ {m} = \sum _ {k=m} ^ {n-1} f _ {k} \Delta g _ {k} + \sum _ {k=m} ^ {n-1} g _ {k} \Delta f _ {k} + \sum _ {k=m} ^ {n-1} \Delta f _ {k} \Delta g _ {k} $$

which is analogous to the integration by parts formula for semimartingales.

Although applications almost always deal with convergence of sequences, the statement is purely algebraic and will work in any field. It will also work when one sequence is in a vector space, and the other is in the relevant field of scalars.

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A community project: prove (or disprove) that $\sum_{n\geq 1}\frac{\sin(2^n)}{n}$ is convergent

As the title says, I would like to launch a community project for proving that the series $$\sum_{n\geq 1}\frac{\sin(2^n)}{n}$$ is convergent. An extensive list of considerations follows. The first fact is that the inequality $$…
Jack D'Aurizio
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Can we have a power density but not a natural density?

For $M \subset \mathbb{N}$ (in this post I follow the convention $\min \mathbb{N} = 1$) and $\alpha \in [0,1]$ define $$S_{M,\alpha}(x) = \sum_{\substack{n\in M \\ n \leqslant x}} \frac{1}{n^{\alpha}}$$ on $[1,+\infty)$. For sets $A \subset B…
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On twisted Euler sums

An interesting investigation started here and it showed that $$ \sum_{k\geq 1}\left(\zeta(m)-H_{k}^{(m)}\right)^2 $$ has a closed form in terms of values of the Riemann $\zeta$ function for any integer $m\geq 2$. I was starting to study the cubic…
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A peculiar Euler sum

I would like a hand in the computation of the following Euler sum (Why isn't here a tag for Euler sums?) $$ S=\sum_{m,n\geq 0}\frac{(-1)^{m+n}}{(2m+1)(2n+1)^2(2m+2n+1)} \tag{1}$$ which arises from the computation of…
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Consider a polygonal line $P_0P_1\ldots P_n$ such that $\angle P_0P_1P_2=\angle P_1P_2P_3=\cdots=\angle P_{n-2}P_{n-1}P_{n}$, all measured clockwise. If $P_0P_1>P_1P_2>\cdots>P_{n-1}P_{n}$, $P_0$ and $P_n$ cannot coincide. While the result is…
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Striking applications of summation by parts

In the same vein as this question Striking applications of integration by parts I'd also like to have a list of some good applications of the discrete version: summation by parts.
user908123
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Summation by Parts

I am trying to solve $\sum\limits_{k=1}^n\frac{2k+1}{k(k+1)}$ using summation by parts: $\sum u\Delta v=uv-\sum Ev\Delta u$ $u = 2x+1, \Delta v=1/x(x+1)=(x-1)_{-2}$ $v=-(x-1)_{-1}=-1/x$ $\Delta u = 2, Ev = -1/(x+1)$ $\sum\frac{(2x+1)}{x(x+1)}\delta…
dbyrne
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If for any $k$, $\sum\limits_{n=0}^\infty a_n^k=\sum\limits_{n=0}^\infty b_n^k$, then$(a_n)=(b_{σ(n)}),\ σ \in{\mathfrak S}_{\mathbb N}$

Let $(a_n)_{n≥0}$ and $(b_n)_{n≥0}$ be two sequences of a nomed algebra such that $\sum{\| a_n\|}$ and $\sum{\| b_n\|}$ converge, and$ $$$\forall n, \ a_n, b_n \neq 0$$ Show that $(\forall k \in \mathbb N^*, \ \sum_{n=0}^{\infty}{a_n}^k =…
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Summation by parts as a case of integration by parts

I was wondering if the summation by parts formula $$ f_ng_n - f_mg_m = \sum_{k=m}^{n-1} f_k(g_{k+1}-g_k) + \sum_{k=m}^{n-1}(f_{k+1}-f_k)g_{k+1} $$ could be proven as a consequence of integration by parts over a suitable domain $$ \int_{[m,n]} d(fg)…
Kolja
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How to show $\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}$ using series manipulation?

Using integration, I managed to show that $$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}\tag1$$ But I would like to prove the equality using series manipulation. Here is what I did:…
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challenging sum $\sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}$

How to prove that \begin{align} \sum_{k=1}^\infty\frac{H_k^{(2)}}{(2k+1)^2}=\frac13\ln^42-2\ln^22\zeta(2)+7\ln2\zeta(3)-\frac{121}{16}\zeta(4)+8\operatorname{Li}_4\left(\frac12\right) \end{align} where…
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Estimating sums over primes

I’m interested in estimating the sum $$ \sum_p \pi\left(\frac{x}{p^3}\right) $$ where the sum is over all primes $p$. (It’s a finite sum of course, you can cut off at $\sqrt[3]{x/2}$.) The goal is a decent estimate, ideally the leading asymptotic…
Charles
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Two formulations of summation by parts

In our lecture we have two different ways to define summation by parts: $$1.)~~\sum\limits_{k=0}^na_k(b_{k+1}-b_k)=a_nb_{n+1}-a_0b_0…
Philipp
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re-calculating exponential sum when exponent changes

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