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Precalculus 8 Online
OpenStudy (anonymous):

Ratio testing 2+4/2^2+8/3^2+16/4^2+......

OpenStudy (anonymous):

2+32/2^2 i'm guessing

OpenStudy (anonymous):

thats not an aswer, i need to know if its divergent or convergent :[

OpenStudy (anonymous):

\[\frac{2^1}{1^2}+\frac{2^2}{2^2}+\frac{2^3}{3^2}+\cdots=\sum_{n=1}^\infty \frac{2^n}{n^2}\] Applying the ratio test, you must figure out the following limit: \[\lim_{n\to\infty}\frac{2^{n+1}}{(n+1)^2}\cdot\frac{n^2}{2^n}\] If the limit is less than 1, the series converges. If the limit is greater than 1, the series diverges. If the limit is equal to 1, the ratio test fails.

OpenStudy (anonymous):

why did you multiply by n^2/2^n ?

OpenStudy (anonymous):

Letting \(a_n\) be the sequence, the ratio test requires that you find the following limit: \[\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\] Here, \(a_{n+1}=\dfrac{2^{n+1}}{(n+1)^2}\) and \(a_n=\dfrac{2^n}{n^2}\). So, \[\frac{a_{n+1}}{a_n}=\frac{\frac{2^{n+1}}{(n+1)^2}}{\frac{2^n}{n^2}}=\frac{2^{n+1}}{(n+1)^2}\cdot\frac{n^2}{2^n}\]

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