Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 :)

share|cite|improve this question
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
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 $$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{1}{3}.$$

share|cite|improve this answer

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

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

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

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


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

share|cite|improve this answer
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:

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.