1. ## Proof by induction

Hi,

$\rm{1.\ Prove\ by\ induction\ that\ 3^n \geq 1 + 2^n\ for\ all\ positive\ integers\ n.}$
A hint or two may suffice with this one.

$\rm{2.\ Show\ that\ all\ arithmetic\ progressions\ are\ divergent.}$

For any arithmetic progression, $\displaystyle S_n = \frac{n}{2}(2a + (n - 1)d),$ which always approaches infinity as $n \rightarrow \infty$.
Therefore any AP is divergent.

2. 1.

Show P(1) is true

$P(1): \ 3\geq 3$

Assume P(k) is true and let $n\leq k$ where k is a fixed arbitrary integer.

$P(k): \ 3^k\geq 1+2^k$

Show P(k+1) is true

Hint: multiple P(k) by 3 and and remember positive terms can be dropped from the inequality.

3. Originally Posted by dwsmith
$P(1): \ 3\geq 3$

Assume P(k) is true and let $n\leq k$ where k is a fixed arbitrary integer.
$P(k): \ 3^k\geq 1+2^k$

Show P(k+1) is true
$3^{k + 1} \geq 3 + 3 \cdot 2^k$
$3^{k + 1} \geq 3 \cdot 2^k$

Is that it?

4. Originally Posted by Hellbent
$3^{k + 1} \geq 3 + 3 \cdot 2^k$
$3^{k + 1} \geq 3 \cdot 2^k$

Is that it?
$3\cdot 3^k\geq 3(1+2^k)\Rightarrow 3^{k+1}\geq 3+(2+1)2^k\Rightarrow 3^{k+1}\geq 1+2+2^{k+1}+2^k$

$k>0$

What can we do to the inequality since k is positive now?

5. $\Rightarrow 3^{k+1}\geq 1+2+2^{k+1}+2^k$
How did you arrive at that?

Originally Posted by dwsmith
What can we do to the inequality since k is positive now?
I do not know.

6. Originally Posted by Hellbent
How did you arrive at that?

I do not know.
$3\cdot 3^k\geq 3\cdot(1+2^k)\Rightarrow 3^{k+1}\geq 3+3\cdot 2^k\Rightarrow 3^{k+1}\geq (2+1)+(2+1)2^k$

7. Originally Posted by Hellbent
I do not know.
You can drop positive values from the inequality.

8. $3^{k+1}\geq 1+2+2^{k+1}+2^k\Rightarrow 3^{k+1}\geq 1+2^{k+1}+(2+2^k)$

$(2+2^k)>0$

You can drop them from the inequality.

9. Originally Posted by Hellbent
Hi,

$\rm{1.\ Prove\ by\ induction\ that\ 3^n \geq 1 + 2^n\ for\ all\ positive\ integers\ n.}$
A hint or two may suffice with this one.

$\rm{2.\ Show\ that\ all\ arithmetic\ progressions\ are\ divergent.}$
For any arithmetic progression, $\displaystyle S_n = \frac{n}{2}(2a + (n - 1)d),$ which always approaches infinity as $n \rightarrow \infty$.
Therefore any AP is divergent.
The problem asked you to show that any arithmetic progression is divergent. You have shown that the series formed by that progression is divergent, not the progression itself.

10. Originally Posted by Hellbent
Hi,

$\rm{1.\ Prove\ by\ induction\ that\ 3^n \geq 1 + 2^n\ for\ all\ positive\ integers\ n.}$
A hint or two may suffice with this one.

P(k)

$3^k\ \ge\ 1+2^k\;\;?$

P(k+1)

$3^{k+1}\ \ge\ 1+2^{k+1}\;\;?$

Try to show that P(k+1) will definately be true "if" P(k) is true.

Therefore, write P(k+1) in terms of P(k) in order to draw a comparison.

Proof

"If" P(k) is true, then

$3^k\ \ge\ 1+2^k$

$\Rightarrow\ 3\left(3^k\right)\ \ge\ 3\left(1+2^k\right)$

$\Rightarrow\ 3^{k+1}\ \ge\ 3+(3)2^k$

$\Rightarrow\ 3^{k+1}\ \ge\ 1+2+(2)2^k+2^k$

$\Rightarrow\ 3^{k+1}\ \ge\ 1+2^{k+1}+\left(2+2^k\right)$

Hence, if P(k) is true, then P(k+1) is true, since $2+2^k>0$

Now all you need do is prove P(k) is true for the first relevant value of "n".

11. Originally Posted by HallsofIvy
The problem asked you to show that any arithmetic progression is divergent. You have shown that the series formed by that progression is divergent, not the progression itself.
$S_{n} = \frac{1}{2}(2a + (n - 1)d)$
with finite values for $a$ and $d$, as $n$ increases, so does the value of $S_n$.
if $n \rightarrow \infty$, then $S_n \rightarrow \infty$ in a positive or negative sense depending on the series.

12. You are showing that the "sum" of the terms, which is the "series", does not converge (your formula is missing an "n").

You are asked to show that the "sequence", the list of terms of $T_n$ or $U_n$, doesn't converge.

$U_n=a+(n-1)d$

13. Originally Posted by Archie Meade
You are showing that the "sum" of the terms, which is the "series", does not converge (your formula is missing an "n").

You are asked to show that the "sequence", the list of terms of $T_n$ or $U_n$, doesn't converge.

$U_n=a+(n-1)d$
$U_{n} = a + (n - 1)d$
with finite values for $a$ and $d$, as $n$ increases, so does the value of $U_n$.
if $n \rightarrow \infty$, then $U_n \rightarrow \infty$ in a positive or negative sense depending on the series.

14. Yes,

to be mathematically concise, you could write

$U_n=a+nd-d=(a-d)+nd=n\left(\frac{a-d}{n}+d\right)$

$n\rightarrow\infty\Rightarrow\ U_n\rightarrow\ nd$

$\Rightarrow\ U_n\rightarrow\pm\infty$