# Thread: how to prove the Abel's test for uniform convergence?

1. ## how to prove the Abel's test for uniform convergence?

The Abel's test for uniform convergence states: Assume $\sum f_n(x)$ converges uniformly on E. A uniformly bounded sequence of functions $\{g_n(x)\}$ satisfies $g_{n+1}(x)\leq g_n(x)$ for all n and x in E. Then $\sum f_n(x)g_n(x)$ converges uniformly.
My tentative proof is as follows:
Let $s_n(x)=\sum\limits_{k = 1}^n f_k(x)g_k(x), F_n(x)=\sum\limits_{k = 1}^n f_k(x), x\in E$. By the partial summation formula, we have $s_n(x)=F_n(x)g_1(x)+\sum\limits_{k = 1}^n (F_n(x)-F_k(x))(g_{k+1}(x)-g_k(x))$. So if n>m we can get $s_n(x)-s_m(x)=(F_n(x)-F_m(x))g_1(x)+\sum\limits_{k = m+1}^n (F_n(x)-F_k(x))(g_{k+1}(x)-g_k(x))$. If the uniform bound for $\{g_n(x)\}$ is M, the absolute value of the above difference satisfies $|s_n(x)-s_m(x)|\leq M|F_n(x)-F_m(x)|+2M\sum\limits_{k = m+1}^n |F_n(x)-F_k(x)|$. From the fact that $\sum f_n(x)$ converges uniformly, for any given $\epsilon>0$, there is an N such that for all m,n>N we have $|F_n(x)-F_m(x)|<\epsilon$. So the right side of the above inequality $\leq [1+2(n-m)]M\epsilon$. But the variable (n-m) makes the Cauchy condition inapplicable. Can you tell me how to proceed, or is there any other better proof? Thanks!

2. the absolute difference should go as follows:
For $N $\sum\limits_{k = m+1}^n |F_n(x)-F_k(x)|[g_k(x)-g_{k+1}(x)]\leq \epsilon(M+\sum\limits_{k = m+1}^n [g_k(x)-g_{k+1}(x)])$ $=\epsilon(M+g_{m+1}(x)-g_{n+1}(x))\leq 3M\epsilon$. Now the Cauchy condition can be applied.
Another proof can be found in K. Knopp's "Theory and application of infinite series", P346-347. This proof is more complicated than the above one.

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# proof of abel's test

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