If each fn is continuous on [a, b] and (sum of)fn converges uniformly on [a, b] then (sum of)Mn

converges, where Mn = max|fn(x)|for x belongs to [a,b].

Is it true?

Thx in advance

.

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- Mar 31st 2009, 11:30 AMwheatkcregarding to Weierstrass M-test
If each fn is continuous on [a, b] and (sum of)fn converges uniformly on [a, b] then (sum of)Mn

converges, where Mn = max|fn(x)|for x belongs to [a,b].

Is it true?

Thx in advance

. - Mar 31st 2009, 12:26 PMOpalg
No, it's not true. But to find a counterexample you have to use some test other than the M-test to prove the uniform convergence. That probably means using the uniform version of the Dirichlet or Abel tests for convergence. A typical counterexample is the series $\displaystyle \sum\frac{(-x)^n}n$ on the interval [0,1]. This converges uniformly by the uniform Dirichlet test. But $\displaystyle \max\{|(-x)^n/n|:x\in[0,1]\} = 1/n$ and $\displaystyle \sum 1/n$ does not converge.

- Mar 31st 2009, 07:08 PMwheatkc
thx...