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I have to determine whether the above sequence is bounded (from above or below).

Bounded from above means: $\frac{2n-1}{3n+1} \le M$ $\forall n$

Bounded from below means: $\frac{2n-1}{3n+1} \ge m$ $\forall n$

Then I tried to solve the above inequalities in n, but they are really complex and it is difficult to interpret the results.

How would you prove that the sequence is bounded?

collimarco
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  • I think proving directly from the inequalities is not so hard. First, divide by $n$ to guess that the fraction $\frac{2-\frac1n}{3+\frac1n} \to \frac23$ as $n\to\infty$, so the upper bound (note that $\frac{2n-1}{3n+1} < \frac{2n}{3n}$) should be $M = \frac23$, and the lower bound should be the first value (I guess you start at $n=1$, which gives $m=\frac14$). So with $M = \frac23$, we have $\frac{2n - 1}{3n+1} \le \frac23$ is same as $3(2n-1) \le 2(3n+1)$ or $-3 \le 2$, which is true. Similarly with $m = \frac14$, we want $4(2n-1) \ge 3n+1$ or $5n \ge 5$, which is true. – ShreevatsaR Jan 13 '14 at 18:23

4 Answers4

4

First prove that it is convergent (Hint: divide by $n$), and then prove that every convergent sequence is bounded.

Prahlad Vaidyanathan
  • 31,554
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If you divide both the top and bottom by $n$, you get a convergent sequence which is therefore bounded. Even more directly, the sequence is clearly positive for all $n$, so you get one bound from below. For an upper bound, observe that $2- \frac{1}{n} <2$ and $3+\frac{1}{n} > 3$ for every $n \in \mathbb{N}.$

B. Mackey
  • 873
3

Hint: Notice that $\displaystyle \frac{2n-1}{3n+1}= \frac{2}{3} \left(1 - \frac{5}{6n+2} \right)$.

Seirios
  • 33,157
1

You could start by asking whether the sequence is monotonic. Calculate the first few values in the sequence to see whether it seems to be increasing or decreasing. Then prove via induction.

If it is monotonic, you'll be able to find an upper or lower bound from the first value. After that, consider whether the sequence seems to be headed toward a specific value given the coefficients on each $n$. Is $\frac{2n-1}{3n}$ larger or smaller than $\frac{2n}{3n}$?

Edit: Actually, you don't need induction for the first part. Instead, compute $\frac{2(n+1) - 1}{3(n+1)+1} - \frac{2n-1}{3n+1}$.

dmk
  • 2,228