# Math Help - 1/n alternating signs

1. ## 1/n alternating signs

Hello again. I'm stuck on how to word my proof exactly or put it in notation or to find out if what I'm writing passes as proof. I'm pretty sure that my proof is right only I think I'm looking for guidelines on how to present it so that it's more readable and easier to understand.

Could you guys critique my proof writing?

Here is the problem:

$\displaystyle s_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \cdots +(-1)^{n+1}\frac{1}{n}$

Show that

$\dfrac{1}{2}\leq s_n \leq 1$ for all n.

Showing the upper bound is easy because for beyond $n=1$ there is always a negative fraction with more magnitude for every positive fraction and in the case that $n$ is odd the left fraction is not enough to make up the gap to push the sum above 1.

To prove the case for the lower term assume again first that $n$ is even. There are then couplets of terms after $-\frac{1}{2}$ of a greater positive fraction and a lesser negative fraction. That progression can only increase after $n=2$ therefore it is bounded below. In the case that n is odd there will simply be a positive fraction added therefore it does not affect our proof of a lower bound.

2. I think your argument is OK.

Here is how I would put the first part. We prove that for all $n\ge 1$, $s_{2n}< 1$ and $s_{2n+1}< 1$ by induction on $n$.

For $n = 1$, $s_2< 1$ and $s_3< 1$. Suppose now that $s_{2n}< 1$ and $s_{2n+1}< 1$, and we need to show that $s_{2n+2}< 1$ and $s_{2n+3}< 1$. Then $s_{2n+2}=s_{2n+1}-1/(2n+2)<1$ and $s_{2n+3}=s_{2n+1}-1/(2n+2)+1/(2n+3)<1$ because $1/(2n+2) > 1/(2n+3)$.

3. Originally Posted by magus
Hello again. I'm stuck on how to word my proof exactly or put it in notation or to find out if what I'm writing passes as proof. I'm pretty sure that my proof is right only I think I'm looking for guidelines on how to present it so that it's more readable and easier to understand.

Could you guys critique my proof writing?

Here is the problem:

$\displaystyle s_n=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4} \cdots +(-1)^{n+1}\frac{1}{n}$

Show that

$\dfrac{1}{2}\leq s_n \leq 1$ for all n.

Showing the upper bound is easy because for beyond $n=1$ there is always a negative fraction with more magnitude for every positive fraction and in the case that $n$ is odd the left fraction is not enough to make up the gap to push the sum above 1.

To prove the case for the lower term assume again first that $n$ is even. There are then couplets of terms after $-\frac{1}{2}$ of a greater positive fraction and a lesser negative fraction. That progression can only increase after $n=2$ therefore it is bounded below. In the case that n is odd there will simply be a positive fraction added therefore it does not affect our proof of a lower bound.
Here's another way to approach this....

$\displaystyle\ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+........$

$=\displaystyle\ 1+\left(\frac{1}{2}-1\right)+\frac{1}{3}+\left(\frac{1}{4}-\frac{1}{2}\right)+\frac{1}{5}+\left(\frac{1}{6}-\frac{1}{3}\right)+\frac{1}{7}+\left(\frac{1}{8}-\frac{1}{4}\right)+........$

$=\displaystyle\ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ \frac{1}{6}+\frac{1}{7}+\frac{1}{8}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}....$

If n=8,

$=\displaystyle\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+ \frac{1}{8}$

which is $<\displaystyle\frac{1}{5}+\frac{1}{5}+\frac{1}{5}+ \frac{1}{5}<1$

and is $>\displaystyle\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+ \frac{1}{8}\ \Rightarrow\ >\frac{1}{2}$

For n even, and greater than 2, we obtain

$S_n=\displaystyle\frac{1}{\frac{n}{2}+1}+\frac{1}{ \frac{n}{2}+2}+....+\frac{1}{n}$

which is $<\displaystyle\left(\frac{1}{\frac{n}{2}+1}\right) \left(\frac{n}{2}\right)$

and $>\displaystyle\frac{1}{n}\ \frac{n}{2}$

$\displaystyle\frac{1}{2}

When n=1, the sum is 1
When n=2, the sum is 0.5