Difference between revisions of "2006 iTest Problems/Problem 31"
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<cmath>\sum_{n=2}^\infty\dfrac{1}{n^2 - 1} + \dfrac{n}{n^4 + 2n^2 + 1 - n^2}</cmath> | <cmath>\sum_{n=2}^\infty\dfrac{1}{n^2 - 1} + \dfrac{n}{n^4 + 2n^2 + 1 - n^2}</cmath> | ||
<cmath>\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)} + \dfrac{n}{(n^2 + n + 1)(n^2 - n + 1)}</cmath> | <cmath>\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)} + \dfrac{n}{(n^2 + n + 1)(n^2 - n + 1)}</cmath> | ||
− | In the fraction <math>\frac{1}{(n+1)(n-1)}</math>, we can split the fraction as <math>\frac{1}{2(n-1)} - \frac{1}{2(n+1)}</math>, making <math>\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)}</math> a [[telescoping]] series. Plugging in values results in <math>\frac12 - \frac16 + \frac14 - \frac18 + \frac16 - \ | + | In the fraction <math>\frac{1}{(n+1)(n-1)}</math>, we can split the fraction as <math>\frac{1}{2(n-1)} - \frac{1}{2(n+1)}</math>, making <math>\sum_{n=2}^\infty\dfrac{1}{(n+1)(n-1)}</math> a [[telescoping]] series. Plugging in values results in <math>\frac12 - \frac16 + \frac14 - \frac18 + \frac16 - \frac1{10} + \cdots = \frac12 + \frac14 = \frac34</math>. |
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Latest revision as of 20:47, 11 December 2018
Problem
The value of the infinite series can be expressed as
where
and
are relatively prime positive numbers. Compute
.
Solution
Notice that , and notice that the numerator contains
. Also, notice that
. We can rewrite the expression by factoring and splitting the fractions.
In the fraction
, we can split the fraction as
, making
a telescoping series. Plugging in values results in
.
In the fraction , plugging in the first few values results in
. The structure suggests that we could try to model the same fraction with a telescoping series.
Substituting for
in
results in
. Also, because
, we find that the expression is equal to
, confirming that
is a telescoping series. The expression equals
.
Therefore, , so
.
See Also
2006 iTest (Problems, Answer Key) | ||
Preceded by: Problem 30 |
Followed by: Problem 32 | |
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