Summation by parts

"Abel transformation" redirects here. For another transformation, see Abel transform.

In mathematics, 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.

Statement

Suppose $\{f_k\}$ and $\{g_k\}$ are two sequences. Then,

\sum_{k=m}^n f_k(g_{k+1}-g_k) = \left[f_{n+1}g_{n+1} - f_m g_m\right] - \sum_{k=m}^n g_{k+1}(f_{k+1}- f_k).

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+1} g_{n+1} - f_m g_m\right] - \sum_{k=m}^n g_{k+1}\Delta f_k,

Note that summation by parts is an analogue to the integration by parts formula,

\int f\,dg = f g - \int g\,df.

Note also that 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.

Newton series

The formula is sometimes given in one of these - slightly different - forms

\begin{align} \sum_{k=0}^n f_k g_k &= f_0 \sum_{k=0}^n g_k+ \sum_{j=0}^{n-1} (f_{j+1}-f_j) \sum_{k=j+1}^n g_k\\ &= f_n \sum_{k=0}^n g_k - \sum_{j=0}^{n-1} \left( f_{j+1}- f_j\right) \sum_{k=0}^j g_k, \end{align}

which represent a special case ( $M=1$ ) of the more general rule

\begin{align}\sum_{k=0}^n f_k g_k &= \sum_{i=0}^{M-1} f_0^{(i)} G_{i}^{(i+1)}+ \sum_{j=0}^{n-M} f^{(M)}_{j} G_{j+M}^{(M)}=\\ &= \sum_{i=0}^{M-1} \left( -1 \right)^i f_{n-i}^{(i)} \tilde{G}_{n-i}^{(i+1)}+ \left( -1 \right) ^{M} \sum_{j=0}^{n-M} f_j^{(M)} \tilde{G}_j^{(M)};\end{align}

both result from iterated application of the initial formula. The auxiliary quantities are Newton series:

f_j^{(M)}:= \sum_{k=0}^M \left(-1 \right)^{M-k} {M \choose k} f_{j+k}

and

G_j^{(M)}:= \sum_{k=j}^n {k-j+M-1 \choose M-1} g_k,

\tilde{G}_j^{(M)}:= \sum_{k=0}^j {j-k+M-1 \choose M-1} g_k.

A remarkable, particular ( $M=n+1$ ) result is the noteworthy identity

\sum_{k=0}^n f_k g_k = \sum_{i=0}^n f_0^{(i)} G_i^{(i+1)} = \sum_{i=0}^n (-1)^i f_{n-i}^{(i)} \tilde{G}_{n-i}^{(i+1)}.

Here, ${n \choose k}$ is the binomial coefficient.

Method

For two given sequences $(a_n) \,$ and $(b_n) \,$ , with $n \in \N$ , one wants to study the sum of the following series:
$S_N = \sum_{n=0}^N a_n b_n$

If we define $B_n = \sum_{k=0}^n b_k,$ then for every $n>0, \,$ $b_n = B_n - B_{n-1} \,$ and

S_N = a_0 b_0 + \sum_{n=1}^N a_n (B_n - B_{n-1}),

S_N = a_0 b_0 - a_0 B_0 + a_N B_N + \sum_{n=0}^{N-1} B_n (a_n - a_{n+1}).

Finally $S_N = a_N B_N - \sum_{n=0}^{N-1} B_n (a_{n+1} - a_n).$

This process, called an Abel transformation, can be used to prove several criteria of convergence for $S_N \,$ .

Similarity with an integration by parts

The formula for an integration by parts is $\int_a^b f(x) g'(x)\,dx = \left[ f(x) g(x) \right]_{a}^{b} - \int_a^b f'(x) g(x)\,dx$
Beside the boundary conditions, we notice that the first integral contains two multiplied functions, one which is integrated in the final integral ( $g' \,$ becomes $g \,$ ) and one which is differentiated ( $f \,$ becomes $f' \,$ ).

The process of the Abel transformation is similar, since one of the two initial sequences is summed ( $b_n \,$ becomes $B_n \,$ ) and the other one is differenced ( $a_n \,$ becomes $a_{n+1} - a_n \,$ ).

Applications

It is used to prove Kronecker's lemma, which in turn, is used to prove a version of the strong law of large numbers under variance constraints.

Summation by parts is frequently used to prove Abel's theorem.

If $\sum b_n$ is a convergent series, and $a_n$ a bounded monotone sequence, then $S_N = \sum_{n=0}^N a_n b_n$ remains a convergent series.

The Cauchy criterion gives

\begin{align}S_M - S_N &= a_M B_M - a_N B_N + \sum_{n=N}^{M-1} B_n (a_{n+1} - a_n)\\&= (a_M-a) B_M - (a_N-a) B_N + a(B_M - B_N) + \sum_{n=N}^{M-1} B_n (a_{n+1} - a_n),\end{align}

where a is the limit of $a_n$ . As $\sum b_n$ is convergent, $B_N$ is bounded independently of $N$ , say by $B$ . As $a_n-a$ go to zero, so go the first two terms. The third term goes to zero by the Cauchy criterion for $\sum b_n$ . The remaining sum is bounded by

\sum_{n=N}^{M-1} |B_n| |a_{n+1}-a_n| \le B \sum_{n=N}^{M-1} |a_{n+1}-a_n| = B|a_N - a_M|

by the monotonicity of $a_n$ , and also goes to zero as $N \to \infty$ .

Using the same proof as above, one shows that

if the partial sums $B_N$ form a bounded sequence independently of $N$ ;
if $\sum_{n=0}^\infty |a_{n+1} - a_n| < \infty$ (so that the sum $\sum_{n=N}^{M-1} |a_{n+1}-a_n|$ goes to zero as $N$ goes to infinity) ; and
if $\lim a_n = 0$

then $S_N = \sum_{n=0}^N a_n b_n$ is a convergent series.

In both cases, the sum of the series satisfies: $|S| = \left|\sum_{n=0}^\infty a_n b_n \right| \le B \sum_{n=0}^\infty |a_{n+1}-a_n|$

References

Abel's lemma at PlanetMath.org.

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