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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 :)

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2  
A decreasing series does not necessarily mean its sum is convergent, For example, $\sum_{n}\frac{1}{n}$. –  Patrick Li Oct 15 '12 at 17:31
    
yes, indeed. But what I used is a criteria - I think so :) (D'Alambert) –  Iuli Oct 15 '12 at 17:33
    
The fact that $\frac{a_{n+1}}{a_n} \leq 1$ is equivalent to $a_{n+1} \leq a_n$ (when the numbers are positive), which is not enough to establish convergence. –  sdcvvc Oct 15 '12 at 17:45
    
@PatrickLi You mention a border case of the ratio test. Given an infinite series $a_0 + a_1 + a_2 + \cdots$, the series is absolutely convergent if the limit of $|a_{n+1}/a_n|$ tends to a limit strictly less than one. The case of $1 + 1/2 + 1/3 + \cdots$ has the limit tending to exactly one. In some sense, almost all decreasing series converge. (Give the space of series a topology inherited from $[0,1]$, where a series is sent to the limit $|a_{n+1}/a_n|$.) –  Fly by Night Oct 15 '12 at 17:57
    
@FlybyNight Are you sure this comment of yours should be addressed to Patrick Li? –  Did Oct 15 '12 at 19:16

3 Answers 3

up vote 5 down vote accepted

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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1  
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. –  Martino Oct 16 '12 at 8:42

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:

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

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