Question

Q: Show that the following sequences are not convergent.
(a) (2n), (b)((-1)nn2)

Answer 1

Answer on Question #64525 – Math – Real Analysis

Question

Show that the following sequences are not convergent:

(a) $2^n$, (b) $(-1)^n n^2$.

Solution

(a) Assume the contrary:

$$\lim 2^n = a \text{ exists. (1)}$$

Then $\forall \varepsilon > 0 \ \exists K \in N$ such that

$$\left| 2^n - a \right| < \varepsilon \quad \forall n > K.$$

If we put $\varepsilon = 1$ we have:

$$\left| 2^n - a \right| < 1 \quad \forall n > K.$$

Hence $2^n < a + 1$. It is known that $n < 2^n$, so

$$\begin{array}{l} n < 2^n < a + 1, \quad (2) \\ n > K. \end{array}$$

Since $a + 1 \in R$, by the Archimedean Property, $\exists M \in N$ such that $M > a + 1$.

Then $\forall n > \max(K, M)$ one gets

$$n > a + 1. \quad (3)$$

There is a contradiction between (2) and (3). It means that assumption (1) is false.

Hence $2^n$ is not convergent.

(b) Assume the contrary:

$$\lim (-1)^n n^2 = a \text{ exists.} \quad (4)$$

Then $\forall \varepsilon > 0 \ \exists K \in N$ such that $|(-1)^n n^2 - a| < \varepsilon \quad \forall n > K$.

If we put $\varepsilon = \frac{1}{2}$ we have: $|(-1)^n n^2 - a| < \frac{1}{2} \quad \forall n > K$.

In particular, $|(-1)^{2n} 4n^2 - a| < \frac{1}{2}$ and $|(-1)^{2n+1} (2n+1)^2 - a| < \frac{1}{2}$.

So $|4n^2 - a| < \frac{1}{2}$ and $|4n^2 + 4n + 1 + a| < \frac{1}{2}$, $\forall n > K$.

Hence

$$\left| 4n^2 - a \right| + \left| 4n^2 + 4n + 1 + a \right| < \frac{1}{2} + \frac{1}{2} = 1,$$

That is,

$$\left| 4n^2 - a \right| + \left| 4n^2 + 4n + 1 + a \right| < 1 \quad (5)$$

Using the triangle inequality,

$$\begin{array}{l} \left| 4n^2 - a \right| + \left| 4n^2 + 4n + 1 + a \right| \geq \left| (4n^2 - a) + (4n^2 + 4n + 1 + a) \right| = \\ = \left| 8n^2 + 4n + 1 \right| > 1, \end{array}$$

because $n > K \geq 1$.

That is,

$$\left| 4n^2 - a \right| + \left| 4n^2 + 4n + 1 + a \right| > 1$$

There is a contradiction between (5) and (6).

It means that assumption (4) is false.

Hence $(-1)^n n^2$ is not convergent.

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