Limit

Calculus Level pending

$If\quad y=\lim _{ n\rightarrow \infty }{ \frac { 1+2+...+n }{ { n }^{ 2 } } } ,\quad then\quad { y }^{ 2 }=?$

The answer is 0.25.

1 solution

Justin Tuazon
Dec 13, 2014

$\quad y=\lim _{ n\rightarrow \infty }{ \frac { 1+2+...+n }{ { n }^{ 2 } } } \\ y=\lim _{ n\rightarrow \infty }{ \frac { \frac { n(n+1) }{ 2 } }{ { n }^{ 2 } } } \\ y=\lim _{ n\rightarrow \infty }{ \frac { n+1 }{ 2n } } \\ y=\lim _{ n\rightarrow \infty }{ \frac { 1+\frac { 1 }{ n } }{ 2 } } \\ y=\frac { 1 }{ 2 } \\ \therefore \boxed { { y }^{ 2 }=\frac { 1 }{ 4 } =0.25 }$

Limit

The answer is 0.25.

1 solution

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