Abel’s Partial Summation Formula

When we consider an infinite series we construct the sequence of partial sums of the series. This is something like the indefinite integral of the sequence of terms of the series.

What’s the analogue of differentiation? We simply take a sequence $A_n$ and write $a_0=A_0$ and $a_n=A_n-A_{n-1}$ for $n\geq1$ . Then we can take the sequence of partial sums

$\displaystyle\sum\limits_{k=0}^na_k=A_0+\sum\limits_{k=1}^n(A_k-A_{k-1})=A_n$

Similarly, we can take the sequence of differences of a sequence of partial sums

$\displaystyle\sum\limits_{k=0}^0a_k=a_0$
$\displaystyle\sum\limits_{k=0}^na_k-\sum\limits_{k=0}^{n-1}a_k=a_n$

This behaves a lot like the Fundamental Theorem of Calculus, in that constructing the sequence of partial sums and constructing the sequence of differences invert each other.

Now how far can we push this analogy? Let’s take two sequences, $a_n$ and $B_n$ . We define the sequence of partial sums $A_n=\sum_{k=0}^na_k$ and the sequence of differences $b_0=B_0$ and $b_n=B_n-B_{n-1}$ . We calculate

$\displaystyle\sum\limits_{k=0}^na_kB_k=A_0B_0+\sum\limits_{k=1}^n(A_k-A_{k-1})B_k=$
$\displaystyle=A_0B_0+\sum\limits_{k=1}^nA_kB_k-\sum\limits_{k=0}^{n-1}A_kB_{k+1}=$
$\displaystyle=\sum\limits_{k=0}^nA_kB_k-\sum\limits_{k=0}^nA_kB_{k+1}+A_nB_{n+1}=$
$\displaystyle=A_nB_{n+1}-\sum\limits_{k=0}^nA_k(B_{k+1}-B_k)=A_nB_{n+1}-\sum\limits_{k=0}^nA_kb_{k+1}$

This is similar to the formula for integration by parts, and is referred to as Abel’s partial summation formula. In particular, it tells us that the series $\sum_{k=0}^\infty a_kB_k$ converges if both the series $\sum_{k=0}^\infty A_kb_{k+1}$ and the sequence $A_nB_{n+1}$ converge.

2 Comments »

There is an interesting application of Abel’s summation to a reformulation of the Riemann Hypothesis in terms of the Mobius function.

Comment by misha | May 1, 2008 | Reply
[…] and Abel’s Tests We can now use Abel’s partial summation formula to establish a couple other convergence […]

Pingback by Dirichlet’s and Abel’s Tests « The Unapologetic Mathematician | May 1, 2008 | Reply

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