Does the formula of the sum of powers help?

Calculus Level 4

$\large \lim_{n\to\infty} \left[n^2 (1^n+2^n+3^n+\ldots+3458^n)^{\frac1n} \left(1-\cos\frac1n\right) \right ]$

Evaluate the limit above.

The answer is 1729.

1 solution

Rohit Ner
Jun 15, 2015

$\displaystyle\lim _{ n\rightarrow \infty }{ { \left[ { \left( { 1 }^{ n }+{ 2 }^{ n }+{ 3 }^{ n }+...+3458^{ n } \right) }^{ \frac { 1 }{ n } }.{ n }^{ 2 }.\left( 1-\cos { \left( \frac { 1 }{ n } \right) } \right) \right] } } \\ = \displaystyle\lim _{ n\rightarrow \infty }{ { \left[ { 3458\left( { \left( \frac { 1 }{ 3458 } \right) }^{ n }+{ \left( \frac { 2 }{ 3458 } \right) }^{ n }+{ \left( \frac { 3 }{ 3458 } \right) }^{ n }+...+{ \left( \frac { 3458 }{ 3458 } \right) }^{ n } \right) }^{ \frac { 1 }{ n } }.\frac { \left( 1-\cos { \left( \frac { 1 }{ n } \right) } \right) }{ { \left( \frac { 1 }{ n } \right) }^{ 2 } } \right] } } \\ = 3458\left( \frac { 1 }{ 2 } \right) \\ \Huge\color{#3D99F6}{=\boxed{1729}}$

Good problem !!! although I do not like the number 1729 that much .

Arghyadeep Chatterjee - 3 years ago

Does the formula of the sum of powers help?

The answer is 1729.

1 solution

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