MHB Determining the convergence or divergence of a sequence using comparison test

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The series $$\sum_{k = 1}^{\infty} {4}^{\frac{1}{k}}$$ is analyzed using the comparison test with the divergent series $$\sum_{k = 1}^{\infty} {4}^{k}$$. It is established that both series are positive and that if one diverges, it does not guarantee the divergence of the other. The limit of the terms in the first series is calculated as $$\lim_{k \to \infty} 4^{\frac{1}{k}} = 1$$, which is not zero, indicating that the series diverges. Therefore, the conclusion is that the series $$\sum_{k = 1}^{\infty} {4}^{\frac{1}{k}}$$ diverges.
tmt1
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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.
 
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$a_k=1^{1/k},\,b_k=4^{1/k}$, 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, $4^{1/k}>1$.
 
tmt said:
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!
 
Prove It said:
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
 

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