Determine convergence or divergence: $\displaystyle a_n = \frac {1+(-1)^n}{n^2}$ Does the this function converge and is the limit equal to 0? That is what I found it to be.
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Originally Posted by saiyanmx89 Determine convergence or divergence: $\displaystyle a_n = \frac {1+(-1)^n}{n^2}$ Does the this function converge and is the limit equal to 0? That is what I found it to be. yes. one way to see this is to use the squeeze theorem
hi I guess you could do $\displaystyle \frac{1 +(-1)^{n}}{n^{2}} \, \leq \, \frac{(-1)^{n}}{n^{2}} $ And the series to the right is converging by Leibniz convergence theorem, because the series is alternating and decreasing.
Originally Posted by Twig hi I guess you could do $\displaystyle \frac{1 +(-1)^{n}}{n^{2}} \, \leq \, \frac{(-1)^{n}}{n^{2}} $ [snip] This inequality is false for even values of n.
$\displaystyle (-1)^n$ is bounded and $\displaystyle \frac1{n^2}\to0$ as $\displaystyle n\to\infty,$ thus $\displaystyle \frac{(-1)^n}{n^2}\to0$ as $\displaystyle n\to\infty,$ and the conclusion follows.
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