Question #111261

Use Weierstrass’ M-Test to prove that the series ∑(x^nXn^2)

∞

n=1


n converges uniformly in x

the interval .

(0,1/5)

Expert's answer

EXPLANATION

For all x∈(0,15)x\in \left( 0,\frac { 1 }{ 5 } \right) and n≥1n\ge 1\quad

0<n2xn<n25n0<{ n }^{ 2 }{ x }^{ n }<\frac { { n }^{ 2 } }{ { 5 }^{ n } } (1)

lim⁡n ⇢∞n25nn=lim⁡n⇢∞(nn)25=15<1\lim _{ n\ \dashrightarrow \infty }{ \sqrt [ n ]{ \frac { { n }^{ 2 } }{ { 5 }^{ n } } } } =\lim _{ n\dashrightarrow \infty }{ \frac { { \left( \sqrt [ n ]{ n } \right) }^{ 2 } }{ 5 } =\frac { 1 }{ 5 } <1\quad }(lim⁡n→∞nn=1)\left( \lim _{ n\rightarrow \infty }{ \sqrt [ n ]{ n } =1 } \right)

Consequently , by Cauchy root test the series

∑n=1∞n25n \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 5 }^{ n } } \quad \quad \quad \ } (2)

converges.

(1) and the convergence of series (2) means that the conditions of Weierstrass’ M-Test are satisfied. Therefore, the series ∑n=1∞n2xn\sum _{ n=1 }^{ \infty }{ { n }^{ 2 } } { x }^{ n }\quadconverges uniformly on the set (0,15)\left( 0,\frac { 1 }{ 5 } \right)


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