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I am trying to prove Abel's Test

Abel's Test: Let $f_n(x)$ be a non-increasing sequence of functions such that $0 \le f_n(x) \le M$ for all $x \in [a,b]$. If $\sum a_n$ converges then $\sum a_nf_n(x)$ converges uniformly in $[a,b]$.

What I tried to do:

Let $A_j= \sum_{n=1}^ja_n$

Then, using summation by parts:

$s_m(x)=\sum_{n=1}^m a_nf_n(x) = \sum_{n=1}^m ((A_n-A_m)(f_n(x)-f_{n+1}(x))) + A_mf_1(x)$

Then, if $m>k$

$$s_m(x)-s_k(x) = f_1(x)(A_m-A_k) + \sum_{n=1}^m((A_n-A_m)(f_n(x)-f_{n+1}(x))) - \sum_{n=1}^k((A_n-A_k)(f_n(x)-f_{n+1}(x)))$$

Then, $|s_m(x)-s_k(x)|\le$

$$|f_1(x)||A_m-A_k|+\sum_{n=1}^m(|A_n-A_m||f_n(x)-f_{n+1}(x)|) - \sum_{n=1}^k(|A_n-A_k||f_n(x)-f_{n+1}(x)|)$$

Since $\sum a_n$ converges, given $\epsilon >0$, exists $N \in \mathbb{N}$ such that $m>k>N \Rightarrow |A_m-A_k|<\epsilon$

Hence,$|s_m(x)-s_k(x)|\le$ $$M\epsilon+\sum_{n=1}^m M|A_n-A_m| - \sum_{n=1}^kM|A_n-A_k|$$

My problem is to bound $|A_n-A_m|$ and $|A_n-A_k|$, since I can not guarantee that $n>N$

idk
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1 Answers1

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\begin{align*} &\left|\sum_{n=1}^{k}(A_{n}-A_{m})(f_{n}(x)-f_{n+1}(x))-\sum_{n=1}^{k}(A_{n}-A_{k})(f_{n}(x)-f_{n+1}(x))\right|\\ &=\left|\sum_{n=1}^{k}(A_{k}-A_{m})(f_{n}(x)-f_{n+1}(x))\right|\\ &\leq\sum_{n=1}^{k}|A_{k}-A_{m}|(f_{n}(x)-f_{n+1}(x))\\ &=|A_{k}-A_{m}|(f_{1}(x)-f_{k+1}(x))\\ &\leq M|A_{k}-A_{m}|. \end{align*} Meanwhile, \begin{align*} \left|\sum_{n=k+1}^{m}(A_{n}-A_{m})(f_{n}(x)-f_{n+1}(x))\right|&\leq\epsilon\sum_{n=k+1}^{m}(f_{n}(x)-f_{n+1}(x))\\ &=\epsilon(f_{k+1}(x)-f_{m+1}(x))\\ &\leq M\epsilon. \end{align*}

user284331
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  • I did not understand the passage from the first to the second line. And the first term should be $|\sum_{n=1}^m(A_n-A_m)(f_n(x)-f_{n+1}(x)) - \sum_{n=1}^k(A_n-A_k)(f_n(x)-f_{n+1}(x))|$ – idk Mar 30 '18 at 23:11
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    I did split it, I consider from $n=1$ to $n=k$ first, and then I control the rest $n=k+1$ to $n=m$ individually. – user284331 Mar 31 '18 at 01:59