Reciprocal of Squares

Algebra Level 4

Given that $\displaystyle \sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } } \approx 1.645$ ,

If $X=\frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +...+\frac { 1 }{ { 100 }^{ 2 } }$ , find the first 2 digits after the decimal point of X. (Note: the 'first 2 digits' is without rounding.)

The answer is 63.

1 solution

K. J. W.
Mar 3, 2016

$\\ \frac { 1 }{ 100 } =\frac { 1 }{ 100\times 101 } +\frac { 1 }{ 101\times 102 } +...\\ \quad \quad \quad \quad >\frac { 1 }{ { 101 }^{ 2 } } +\frac { 1 }{ { 102 }^{ 2 } } +...\\ \quad \quad \quad \quad >\frac { 1 }{ 101\times 102 } +\frac { 1 }{ 102\times 103 } +...\\ \quad \quad \quad \quad =\frac { 1 }{ 101 } \\ Therefore\\ \frac { 1 }{ { 101 }^{ 2 } } +\frac { 1 }{ { 102 }^{ 2 } } +...\approx 0.0099...\\ \therefore \frac { 1 }{ { 1 }^{ 2 } } +\frac { 1 }{ { 2 }^{ 2 } } +...+\frac { 1 }{ { 100 }^{ 2 } } =\sum _{ k=1 }^{ \infty }{ \frac { 1 }{ { k }^{ 2 } } } -(\frac { 1 }{ { 101 }^{ 2 } } +\frac { 1 }{ { 102 }^{ 2 } } +...)\\ \approx 1.635\quad and\quad the\quad two\quad digits\quad after\quad the\quad decimal\quad point\quad are\quad \boxed { 63 } .$

How you write the first four steps... Till before therefore

Punithan Mech - 5 years, 2 months ago

$\frac { 1 }{ a(a+1) } <\frac { 1 }{ a(a) } <\frac { 1 }{ a(a-1) } \quad for\quad a\ge 2$

Also, $\frac { 1 }{ a(a+1) } =\frac { 1 }{ a } -\frac { 1 }{ a+1 }$ , so the terms cancel out to 1/100 and 1/101 since the sum is to infinity.

K. J. W. - 5 years, 2 months ago

Reciprocal of Squares

The answer is 63.

1 solution

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