Proving Abel's theorem (exercise from Stein complex analysis book)

Question

If we use the hint, consider the sequence $\sum_{k=1}^n r^k a_k $and let $B_k = \sum_{i=1}^k a_i$ . Then according to the summation by parts formula:

$$ \sum_{k=1}^n r^k a_k = r^n B_n - r B_1 - \sum_{k=1}^{n-1} (r^{k+1} - r^k) B_k$$

If we let $r \to 1$ then LHS becomes $B_n - B_1$ and as $n \to \infty$ we obtain $\sum a_n - B_1$, what am I doing wrong here? But, more importantly what confuses me is why do we need summation by parts? Why can't we just apply $r \to 1$ to the sequence of partial sums $\sum^n r^k a_k$ and then let $n \to \infty$? This is what confuses me.

Answer 1

Let $a_0=0$. If $B_n=\sum_{k=0}^na_n$, then, when $|r|<1$,\begin{align}\sum_{n=0}^\infty a_nr^n&=\sum_{n=0}^\infty B_nr^n-\sum_{n=1}^\infty B_{n-1}r^n\\&=\sum_{n=0}^\infty B_nr^n-r\sum_{n=0}^\infty B_nr^N\\&=(1-r)\sum_{n=0}^\infty B_nr^n\end{align}(this is what you would get applying summation by parts). So, if $B=\sum_{n=0}^\infty a_n$, we have$$B=B(1-r)\frac1{1-r}=(1-r)\sum_{n=1}^\infty Br^n$$and therefore, for each $r\in(-1,1)$,$$\left(\sum_{n=0}^\infty a_nr^n\right)-B=(1-r)\sum_{n=0}^\infty(B_n-B)r^n.$$Now, if $\varepsilon>0$, take $N\in\Bbb N$ such that $|B-B_n|<\frac\varepsilon2$ when $n\geqslant N$ and take $\delta>0$ such that$$r\in(1-\delta,1)\implies\left|(1-r)\sum_{n=0}^{N-1}(B_n-B)r^n\right|<\frac\varepsilon2.$$Then, if $r\in(1-\delta,1)$,\begin{align}\left|\left(\sum_{n=0}^\infty a_nr^n\right)-B\right|&=\left|(1-r)\sum_{n=0}^\infty(B_n-B)r^n\right|\\&\leqslant\left|(1-r)\sum_{n=0}^{N-1}(B_n-B)r^n\right|+\left|(1-r)\sum_{n=N}^\infty(B_n-B)r^n\right|\\&<\frac\varepsilon2+(1-r)\sum_{n=N}^\infty\frac\varepsilon2r^n\\&<\frac\varepsilon2+\frac\varepsilon2\\&=\varepsilon.\end{align}

You cannot approach this by simply stating that$$\lim_{r\to1}\sum_{n=1}^\infty a_nr^n=\sum_{n=1}^\infty\lim_{r\to1}a_nr^n=\sum_{n=1}^\infty a_n$$because, in general, a limit and an infinite sum do not commute. For instance,$$\lim_{x\to\infty}\sum_{n=0}^\infty\frac{x^n}{e^xn!}=\lim_{x\to\infty}1=1,$$whereas$$\sum_{n=0}^\infty\lim_{x\to\infty}\frac{x^n}{e^xn!}=\sum_{n=0}^\infty0=0.$$

Answer 2

Your argument is proving $$ \lim_{n\to\infty}\left( \lim_{r\to1}\sum_{k=1}^n r^ka_k\right)=\sum_{n=1}^\infty a_n\tag1 $$ This is true, but is not what Abel's theorem is claiming, which is: $$ \lim_{r\to1}\left( \lim_{n\to\infty} \sum_{k=1}^n r^ka_k\right)=\sum_{n=1}^\infty a_n\tag2 $$ The reason why result (2) is different from (1) is that you are interchanging two limit operations, which is not always legal. To justify (2) you need a more careful argument, like summation by parts.