Think about it!

Calculus Level pending

Find the smallest possible value of $\alpha$ such that;

$\sum _{ n=1 }^{ N }{ \frac { 2n+1 }{ ({ n }^{ 2 }+1)({ n }^{ 2 }+2n+2) } } \le \alpha$

Where $N\ge 1$

Impossible to determine

\frac { 1 }{ 4 }

\frac { 1 }{ 2 }

1 solution

Adil Brohi
Jul 16, 2014

Begin with breaking the expression into partial fractions;

${ \frac { 2n+1 }{ ({ n }^{ 2 }+1)({ n }^{ 2 }+2n+2) } }\equiv \frac { 1 }{ { (n }^{ 2 }+1) } -\frac { 1 }{ { (n+1) }^{ 2 }+1 }$

Now using method of differences;

$\frac { 1 }{ 2 } -\frac { 1 }{ 5 } +\frac { 1 }{ 5 } -\frac { 1 }{ 10 } +....\frac { 1 }{ (N^{ 2 }+1) } -\frac { 1 }{ { (N+1) }^{ 2 }+1 } \\ \sum _{ n=1 }^{ N }{ \frac { 2n+1 }{ ({ n }^{ 2 }+1)({ n }^{ 2 }+2n+2) } =\frac { 1 }{ 2 } -\frac { 1 }{ { (N+1) }^{ 2 }+1 } }$

$Series\quad converges\quad when\quad when\quad N\rightarrow \infty \\ \sum _{ n=1 }^{ N }{ \frac { 2n+1 }{ ({ n }^{ 2 }+1)({ n }^{ 2 }+2n+2) } =\frac { 1 }{ 2 } -\frac { 1 }{ { (N+1) }^{ 2 }+1 } } \\ \sum _{ n=1 }^{ \infty }{ \frac { 2n+1 }{ ({ n }^{ 2 }+1)({ n }^{ 2 }+2n+2) } \le \frac { 1 }{ 2 } } \\ \boxed { \alpha =\frac { 1 }{ 2 } }$

This solution sure rings a bell. See my solution to this one

Simple Infinite Progression

Michael Mendrin - 6 years, 11 months ago

Think about it!

1 solution

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