Determining the convergence or divergence of a sequence using comparison test

[tmt1]

I have this series:

$$\sum_{k = 1}^{\infty} {4}^{\frac{1}{k}}$$

To solve this, I am trying to compare it to this series

$$\sum_{k = 1}^{\infty} {4}^{k}$$

So, I can let $a_k = {4}^{\frac{1}{k}} $ and $b_k = {4}^{k}$

These seem to be both positive series and $ 0 \le a_k \le b_k$

Therefore, if $\sum_{}^{} b_k$ converges then $\sum_{}^{} a_k$ converges and if $\sum_{}^{} a_k$ diverges, then $\sum_{}^{} b_k $ diverges.

However, if $\sum_{}^{} b_k $ diverges, that doesn't guarantee that $\sum_{}^{} a_k$ diverges.

In this case, $\sum_{}^{} b_k $ diverges as $$\sum_{k = 1}^{\infty} {4}^{k}$$ diverges (as $4 > 1$). However, I'm not sure what to conclude with this information.

 

[Greg]

$a_k=1^{1/k},\,b_k=4^{1/k}$, but $\sum a_k$ diverges and $b_k>a_k\forall\,k\in\mathbb{N}$. Without comparison, simply note that no matter how large $k$ gets, $4^{1/k}>1$.

 

[Prove It]

I have this series:

$$\sum_{k = 1}^{\infty} {4}^{\frac{1}{k}}$$

To solve this, I am trying to compare it to this series

$$\sum_{k = 1}^{\infty} {4}^{k}$$

So, I can let $a_k = {4}^{\frac{1}{k}} $ and $b_k = {4}^{k}$

These seem to be both positive series and $ 0 \le a_k \le b_k$

Therefore, if $\sum_{}^{} b_k$ converges then $\sum_{}^{} a_k$ converges and if $\sum_{}^{} a_k$ diverges, then $\sum_{}^{} b_k $ diverges.

However, if $\sum_{}^{} b_k $ diverges, that doesn't guarantee that $\sum_{}^{} a_k$ diverges.

In this case, $\sum_{}^{} b_k $ diverges as $$\sum_{k = 1}^{\infty} {4}^{k}$$ diverges (as $4 > 1$). However, I'm not sure what to conclude with this information.

A series only ever has the possibility of converging if the sequence of values being added decreases to 0. In other words, if $\displaystyle \begin{align*} \lim_{k \to \infty} 4^{\frac{1}{k}} \neq 0 \end{align*}$ then the series diverges.

But hang on, $\displaystyle \begin{align*} \lim_{k \to \infty} 4^{\frac{1}{k}} = 4^0 = 1 \end{align*}$. As this is not zero the series must diverge.

This is ALWAYS the first test you should try for ANY series!

 

[tmt1]

A series only ever has the possibility of converging if the sequence of values being added decreases to 0. In other words, if $\displaystyle \begin{align*} \lim_{k \to \infty} 4^{\frac{1}{k}} \neq 0 \end{align*}$ then the series diverges.

But hang on, $\displaystyle \begin{align*} \lim_{k \to \infty} 4^{\frac{1}{k}} = 4^0 = 1 \end{align*}$. As this is not zero the series must diverge.

This is ALWAYS the first test you should try for ANY series!

Ok this is very helpful, thank you