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How is this series ?

$\sum_{n \geq1}{\frac{2n^2}{3^n}} ?$

How I made: $a_{n}=\frac{2n^2}{3^n}$ and then $\frac{a_{n+1}}{a_n} \leq 1$ so the series is convergence .

Is ok ?

thanks :)

  • 0
    @did Yes, I am sure.2012-10-15

3 Answers 3

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Having ratio $\le 1$ is not enough for convergence. The simplest example is $1+1+1+\cdots$. There are more subtle examples. (By the way, $\dfrac{a_2}{a_1}\gt 1$, though this does not matter.)

We want to show that there is a fixed $b$ with $0\le b\lt 1$ such that for large enough $n$, $\dfrac{a_{n+1}}{a_n}\le b$. It will be enough to show that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{1}{3}.$

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Since $2n^2 < 4\cdot 2^n$, the series is dominated by a geometric series, so it is convergent. Moreover, we have:

$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{x^n}{3^n} = \frac{x}{3-x}, $

$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{n x^n}{3^n} = x\cdot\frac{d}{dx}\left(\frac{x}{3-x}\right) = \frac{3x}{(x-3)^2}$

$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{n^2 x^n}{3^n} = x\cdot\frac{d}{dx}\left(\frac{3x}{(x-3)^2}\right) = \frac{3x(3+x)}{(3-x)^3}$

so:

$\sum_{n\geq 1}\frac{2n^2}{3^n}=2\cdot\frac{3\cdot 4}{8} = 3.$

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    And this argument works for all series with general term $n^k a^n$ where $k \in \mathbb{N}$ and 0 < a < 1 are fixed.2012-10-16
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Not quite.

You need the limit of the ratio $a_{n+1}/a_n$ to be strictly less than one for the ratio test to apply. In symbols, you need:

$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1 \, . $

In your example, you have:

$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{(n+1)^2}{3n^2}\right| = \frac{1}{3} < 1 \, . $

Thus, by the ratio test, you sequence converges. Moreover, one can show that:

$\lim_{k \to \infty} \left(\sum_{n=0}^k \frac{2n^2}{3^n}\right) = 3 \, . $