MHB Which test do we have to apply at each case?

[mathmari]

Hey!

To check the convergence or divergence of series there are the following tests:

• Trivial test
  If a series $\sum a_k$ converges, then the sequence $a_k$ is a zero-sequence.
  With this test we can just prove the divergence, but not in general the convergence of a series. \
• Ratiom test
  If $\lim \sup |\frac{a_{n+1}}{a_n}|<1$, then the series converges. \\
  If $\lim \sup |\frac{a_{n+1}}{a_n}|>1$, then the series diverges. \\
  If $\lim \sup |\frac{a_{n+1}}{a_n}|=1$, then this test doesn't tell anything about the convergence/divergence.
• Root test
  The series $\sum a_k$ converges absolutely, if $\lim_{k\rightarrow \infty}\sqrt[k]{|a_k|}<1$.
  For $\lim_{k\rightarrow \infty}\sqrt[k]{|a_k|}>1$ the divergence of the series is implied.
• Leibniz test
  If $(a_k)$ is a monotone decreasing zero-sequence, then the alternating seriesn$\sum (-1)^ka_k$ converges.
• Comparison test
  If $0 \leq |a_k|\leq b_k$ from some $k_0$ and $\sum b_k$converges, then the series $\sum a_k$ converges.
  If $0 \leq |a_k|\leq b_k$ from some $k_0$ and $\sum a_k$ diverges, then the series $\sum b_k$ diverges.
• Integral test
  let $f : [1, \infty ) \rightarrow [0, \infty )$ be monotone decreasing. then the series $\sum_{k=1}^{\infty} f(k)$ converges iff the integral $\int_1^{\infty} f(x)ds$ exists.

right? (Wondering) How do we know which test we have to apply at the following cases?

1. $\displaystyle{\sum_{k=1}^{\infty}\frac{1}{5^k}}$
2. $\displaystyle{\sum_{n=0}^{\infty}\frac{2^{3n+1}}{9^n}}$
3. $\displaystyle{\sum_{k=2}^{\infty}1,0001^k}$
4. $\displaystyle{\sum_{k=0}^{\infty}\frac{(-1)^k}{a^k}}$

(Wondering)

 

[Prove It]

Hey!

To check the convergence or divergence of series there are the following tests:

• Trivial test
  If a series $\sum a_k$ converges, then the sequence $a_k$ is a zero-sequence.
  With this test we can just prove the divergence, but not in general the convergence of a series. \
• Ratiom test
  If $\lim \sup |\frac{a_{n+1}}{a_n}|<1$, then the series converges. \\
  If $\lim \sup |\frac{a_{n+1}}{a_n}|>1$, then the series diverges. \\
  If $\lim \sup |\frac{a_{n+1}}{a_n}|=1$, then this test doesn't tell anything about the convergence/divergence.
• Root test
  The series $\sum a_k$ converges absolutely, if $\lim_{k\rightarrow \infty}\sqrt[k]{|a_k|}<1$.
  For $\lim_{k\rightarrow \infty}\sqrt[k]{|a_k|}>1$ the divergence of the series is implied.
• Leibniz test
  If $(a_k)$ is a monotone decreasing zero-sequence, then the alternating seriesn$\sum (-1)^ka_k$ converges.
• Comparison test
  If $0 \leq |a_k|\leq b_k$ from some $k_0$ and $\sum b_k$converges, then the series $\sum a_k$ converges.
  If $0 \leq |a_k|\leq b_k$ from some $k_0$ and $\sum a_k$ diverges, then the series $\sum b_k$ diverges.
• Integral test
  let $f : [1, \infty ) \rightarrow [0, \infty )$ be monotone decreasing. then the series $\sum_{k=1}^{\infty} f(k)$ converges iff the integral $\int_1^{\infty} f(x)ds$ exists.

right? (Wondering) How do we know which test we have to apply at the following cases?

1. $\displaystyle{\sum_{k=1}^{\infty}\frac{1}{5^k}}$
2. $\displaystyle{\sum_{n=0}^{\infty}\frac{2^{3n+1}}{9^n}}$
3. $\displaystyle{\sum_{k=2}^{\infty}1,0001^k}$
4. $\displaystyle{\sum_{k=0}^{\infty}\frac{(-1)^k}{a^k}}$

(Wondering)

The first one can be written as $\displaystyle \begin{align*} \sum_{k = 1}^{\infty}{\frac{1}{5^k}} = \sum_{k = 1}^{\infty}{ \left( \frac{1}{5} \right) ^k } \end{align*}$, a geometric series. When does a geometric series converge?

The second one can be written as

$\displaystyle \begin{align*} \sum_{n = 0}^{\infty}{ \frac{2^{3n+1}}{9^n} } &= 2\sum_{n = 0}^{\infty}{\frac{2^{3n}}{9^n}} \\ &= 2\sum_{n = 0}^{\infty}{ \frac{\left( 2^3 \right) ^n}{9^n} } \\ &= 2\sum_{n=0}^{\infty}{ \frac{8^n}{9^n} } \\ &= 2\sum_{n = 0}^{\infty}{ \left( \frac{8}{9} \right) ^n } \end{align*}$

again, reduced to determining if a geometric series converges.The other two are also geometric series.

 

[mathmari]

The first one can be written as $\displaystyle \begin{align*} \sum_{k = 1}^{\infty}{\frac{1}{5^k}} = \sum_{k = 1}^{\infty}{ \left( \frac{1}{5} \right) ^k } \end{align*}$, a geometric series. When does a geometric series converge?

The second one can be written as

$\displaystyle \begin{align*} \sum_{n = 0}^{\infty}{ \frac{2^{3n+1}}{9^n} } &= 2\sum_{n = 0}^{\infty}{\frac{2^{3n}}{9^n}} \\ &= 2\sum_{n = 0}^{\infty}{ \frac{\left( 2^3 \right) ^n}{9^n} } \\ &= 2\sum_{n=0}^{\infty}{ \frac{8^n}{9^n} } \\ &= 2\sum_{n = 0}^{\infty}{ \left( \frac{8}{9} \right) ^n } \end{align*}$

again, reduced to determining if a geometric series converges.The other two are also geometric series.

We have that for $-1<q<1$ the geometric series converges and is equal to $\displaystyle{\sum_{k=0}^{\infty}q^k=\frac{1}{1-q}}$.

So, we have the following:

1. $\displaystyle {\sum_{k = 1}^{\infty}{\frac{1}{5^k}} = \sum_{k = 1}^{\infty}{ \left( \frac{1}{5} \right) ^k } =\sum_{k = 0}^{\infty}{ \left( \frac{1}{5} \right) ^k }-{ \left( \frac{1}{5} \right) ^0 }=\sum_{k = 0}^{\infty}{ \left( \frac{1}{5} \right) ^k }-1\overset{ -1<\frac{1}{5}<1 }{ = } \frac{1}{1-\frac{1}{5}}-1=\frac{1}{\frac{4}{5}}-1=\frac{5}{4}-1=\frac{1}{4}}$
2. $\displaystyle{ \sum_{n = 0}^{\infty}{ \frac{2^{3n+1}}{9^n} } = 2\sum_{n = 0}^{\infty}{ \left( \frac{8}{9} \right) ^n }}\overset{ -1<\frac{8}{9}<1 }{ = }2\frac{1}{1-\frac{8}{9}}=2\frac{1}{\frac{1}{9}}=18$
3. SInce $1,0001>1$ we have the series does not converge, right? (Wondering)
4. $\displaystyle{\sum_{k=0}^{\infty}\frac{(-1)^k}{a^k}=\sum_{k=0}^{\infty}\left (\frac{-1}{a}\right )^k}$
   Since $a>1$ we have that $0<\frac{1}{a}<1 \Rightarrow -1<-\frac{1}{a}<0$, right? (Wondering)
   Therefore, $\displaystyle{\sum_{k=0}^{\infty}\frac{(-1)^k}{a^k}=\frac{1}{1+\frac{1}{a}}=\frac{a}{a+1}}$

 

[Prove It]

We have that for $-1<q<1$ the geometric series converges and is equal to $\displaystyle{\sum_{k=0}^{\infty}q^k=\frac{1}{1-q}}$.

So, we have the following:

1. $\displaystyle {\sum_{k = 1}^{\infty}{\frac{1}{5^k}} = \sum_{k = 1}^{\infty}{ \left( \frac{1}{5} \right) ^k } =\sum_{k = 0}^{\infty}{ \left( \frac{1}{5} \right) ^k }-{ \left( \frac{1}{5} \right) ^0 }=\sum_{k = 0}^{\infty}{ \left( \frac{1}{5} \right) ^k }-1\overset{ -1<\frac{1}{5}<1 }{ = } \frac{1}{1-\frac{1}{5}}-1=\frac{1}{\frac{4}{5}}-1=\frac{5}{4}-1=\frac{1}{4}}$
2. $\displaystyle{ \sum_{n = 0}^{\infty}{ \frac{2^{3n+1}}{9^n} } = 2\sum_{n = 0}^{\infty}{ \left( \frac{8}{9} \right) ^n }}\overset{ -1<\frac{8}{9}<1 }{ = }2\frac{1}{1-\frac{8}{9}}=2\frac{1}{\frac{1}{9}}=18$
3. SInce $1,0001>1$ we have the series does not converge, right? (Wondering)
4. $\displaystyle{\sum_{k=0}^{\infty}\frac{(-1)^k}{a^k}=\sum_{k=0}^{\infty}\left (\frac{-1}{a}\right )^k}$
   Since $a>1$ we have that $0<\frac{1}{a}<1 \Rightarrow -1<-\frac{1}{a}<0$, right? (Wondering)
   Therefore, $\displaystyle{\sum_{k=0}^{\infty}\frac{(-1)^k}{a^k}=\frac{1}{1+\frac{1}{a}}=\frac{a}{a+1}}$

The first three are correct.

For the fourth, how do you know a > 1?

 

[mathmari]

For the fourth, how do you know a > 1?

Ah, this is given. I forgot to mention it in my first post. Sorry. (Blush)

Since $a>1$ we have that $0<\frac{1}{a}<1 \Rightarrow -1<-\frac{1}{a}<0$

This implication is correct, isn't it? (Wondering)

 

[Prove It]

Ah, this is given. I forgot to mention it in my first post. Sorry. (Blush)
This implication is correct, isn't it? (Wondering)

I agree that $\displaystyle \begin{align*} a > 1 \end{align*}$ implies that $\displaystyle \begin{align*} -1 < -\frac{1}{ a} < 0 \end{align*}$, and the geometric series to converge, we also require that $\displaystyle \begin{align*} \left| r \right| = \left| -\frac{1}{a} \right| < 1 \end{align*}$, which we can obviously see it does. So that means the series will converge.

 

[mathmari]

I agree that $\displaystyle \begin{align*} a > 1 \end{align*}$ implies that $\displaystyle \begin{align*} -1 < -\frac{1}{ a} < 0 \end{align*}$, and the geometric series to converge, we also require that $\displaystyle \begin{align*} \left| r \right| = \left| -\frac{1}{a} \right| < 1 \end{align*}$, which we can obviously see it does. So that means the series will converge.

Ok... Thank you very much! (Mmm)