EXPLANATION
For all x ∈ ( 0 , 1 5 ) x\in \left( 0,\frac { 1 }{ 5 } \right) x ∈ ( 0 , 5 1 ) and n ≥ 1 n\ge 1\quad n ≥ 1
0 < n 2 x n < n 2 5 n 0<{ n }^{ 2 }{ x }^{ n }<\frac { { n }^{ 2 } }{ { 5 }^{ n } } 0 < n 2 x n < 5 n n 2 (1)
lim n ⇢ ∞ n 2 5 n n = lim n ⇢ ∞ ( n n ) 2 5 = 1 5 < 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 ⇢ ∞ n 5 n n 2 = lim n ⇢ ∞ 5 ( n n ) 2 = 5 1 < 1 ( lim n → ∞ n n = 1 ) \left( \lim _{ n\rightarrow \infty }{ \sqrt [ n ]{ n } =1 } \right) ( lim n → ∞ n n = 1 )
Consequently , by Cauchy root test the series
∑ n = 1 ∞ n 2 5 n \sum _{ n=1 }^{ \infty }{ \frac { { n }^{ 2 } }{ { 5 }^{ n } } \quad \quad \quad \ } ∑ n = 1 ∞ 5 n n 2 (2)
converges.
(1) and the convergence of series (2) means that the conditions of Weierstrass’ M-Test are satisfied. Therefore, the series ∑ n = 1 ∞ n 2 x n \sum _{ n=1 }^{ \infty }{ { n }^{ 2 } } { x }^{ n }\quad ∑ n = 1 ∞ n 2 x n converges uniformly on the set ( 0 , 1 5 ) \left( 0,\frac { 1 }{ 5 } \right) ( 0 , 5 1 )
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