summation by parts

The following corollaries apply Abel’s lemma to allow estimation of certain bounded sums:

Corollary 1

(Summation by parts)
Let $\mathrm{\{}{a}_i\mathrm{\}}\mathrm{,}\mathrm{\{}{b}_i\mathrm{\}}$ be sequences of complex numbers. Suppose the partial sums of the $a_i$ are bounded in magnitude by $h$, that ${\mathrm{\sum }}_{\mathrm{0}}^{\mathrm{\infty }}\mathrm{|}{b}_i\mathrm{-}{b}_{i\mathrm{+}\mathrm{1}}\mathrm{|}$ converges, and that ${\lim }_{i\mathrm{\to }\mathrm{\infty }}\mathit{}{b}_i\mathrm{=}\mathrm{0}$. Then ${\mathrm{\sum }}_{\mathrm{0}}^{\mathrm{\infty }}{a}_i\mathit{}{b}_i$ converges, and

$$\left|\sum _0^{\mathrm{\infty }}{a}_i{b}_i\right|\le h\sum _0^{\mathrm{\infty }}|b_i-b_{i+1}|$$

Proof. By Abel’s lemma,

$$\sum _{i=0}^N{a}_i{b}_i=\sum _{i=0}^{N-1}{A}_i(b_i-b_{i+1})+A_N{b}_N$$

so that

$\left|{\displaystyle \sum _{i=0}^N}{a}_i{b}_i\right|$$=\left|{\displaystyle \sum _{i=0}^{N-1}}{A}_i(b_i-b_{i+1})+A_N{b}_N\right|\le {\displaystyle \sum _{i=0}^{N-1}}\left|A_i(b_i-b_{i+1})\right|+\left|A_N{b}_N\right|$

$\le h{\displaystyle \sum _{i=0}^{N-1}}\left|b_i-b_{i+1}\right|+h\left|b_N\right|$

The condition that the $b_i\to 0$ is easily seen to imply that the sequence $\left|{\sum }_{i=0}^N{a}_i{b}_i\right|$ is Cauchy hence convergent, so that

$$\left|\sum _{i=0}^{\mathrm{\infty }}{a}_i{b}_i\right|\le h\sum _{i=0}^{\mathrm{\infty }}\left|b_i-b_{i+1}\right|$$

since $b_N\to 0$.

Corollary 2

(Summation by parts for real sequences)
Let $\mathrm{\{}{a}_i\mathrm{\}}$ be a sequence of complex numbers. Suppose the partial sums are bounded in magnitude by $h$. Let $\mathrm{\{}{b}_i\mathrm{\}}$ be a sequence of decreasing positive real numbers such that ${\lim }_{i\mathrm{\to }\mathrm{\infty }}\mathit{}{b}_i\mathrm{=}\mathrm{0}$. Then ${\mathrm{\sum }}_{\mathrm{1}}^{\mathrm{\infty }}{a}_i\mathit{}{b}_i$ converges, and $\mathrm{|}{\mathrm{\sum }}_{\mathrm{1}}^{\mathrm{\infty }}{a}_i\mathit{}{b}_i\mathrm{|}\mathrm{\le }h\mathit{}{b}_{\mathrm{1}}$.

Proof. This follows immediately from the above, since $|b_i-b_{i+1}|=b_i-b_{i+1}$.