Show that the series diverges $1 + \frac{3}{2} +\frac{9}{4} + \frac{27}{8} +\frac{81}{16}+\cdots$

Question

Show that the series diverges $$1 + \frac{3}{2} +\frac{9}{4} + \frac{27}{8} +\frac{81}{16}+\cdots$$

The numerator is increasing at a rate of $3a_n$ and the denominator is simply doubling $2b_n$

Naturally this series will diverge since it will increase to infinity since the numerator is growing faster than the denominator.

Is there a more robust way of illustrating this?

Answer 1

Use the ratio test; the ratio of consecutive terms is $\frac32>1$. You could also use this theorem.

Answer 2

We have

$$1 + \frac{3}{2} +\frac{9}{4} + \frac{27}{8} +\frac{81}{16}+...=\sum_{n=0}^\infty \left(\frac32\right)^n$$

and $a_n \to \infty$ therefore the series diverges, indeed recall that

$$\sum_{n=0}^\infty a_n <\infty \implies a_n \to 0$$

Answer 3

For a series $\sum_{i=1}^\infty a_i$, let $S_n=\sum_{i=1}^n a_i$. Then by definition, the series $\sum_{i=1}^\infty a_i$ converges if and only if the sequence $S_n=\sum_{i=1}^n a_i$ converges. Now we can easily prove the following:

If the sequence of terms $\{a_i\}$ does not converge to $0$, then the series $\sum_{i=1}^\infty a_i$ diverges.

Proof: if $\sum_{i=1}^\infty a_i$ converges to some $L$, then for any $\varepsilon>0$, there exists $N$ such that $|S_n-L|<\varepsilon$ for all $n>N$. In particular, $|a_{n+1}|= |S_{n+1}-S_n|\le |S_{n+1}-L|+|L-S_n|<2\varepsilon$, so $\{a_n\}$ converges to $0$.

Applying this to your example, the series diverges since $\left\{\left(\frac{3}{2}\right)^i\right\}$ does not converge to $0$.

Answer 4

HINT

This is the series: $$\sum_{n=0}^{\infty}\big(\frac{3}{2}\big)^n$$

this is a geometric series, to what sum does the general term converges a n goes to infinity