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Math Help - 1/n alternating signs

  1. #1
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    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.
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  2. #2
    MHF Contributor
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    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).
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  3. #3
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    Quote Originally Posted by magus View Post
    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}<S_n<\frac{2}{n+2}\ \frac{n}{2}

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