Could someone take a look at an analysis proof?

[bennyska]

Homework Statement

first off, my latex isn't coming out just right. I've attached a pdf that should look right. if you can fill in the gaps from what I've posted here, sweet, otherwise, if you wouldn't mind checking out the attached pdf, that would be awesome.
Assume that $\sum_{k=1}^{\infty}a_{k}$ converges with each $a_{k} &gt;0. $<br /> Let $s_{n}=\sum_{k=1}^{\infty}a_{k}.$ Show that $\sum_{k=1}^{\infty}a_{k}/s_{k}$<br /> converges.<br /> <br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> Assume that $\sum_{k=1}^{\infty}a_{k}$ converges with each $a_{k} &amp;gt;0. $&lt;br /&gt; Let $s_{n}=\sum_{k=1}^{\infty}a_{k}.$ Show that $\sum_{k=1}^{\infty}a_{k}/s_{k}$&lt;br /&gt; converges.&lt;br /&gt; &lt;br /&gt; Proof: Since $\sum_{k=1}^{\infty}a_{k}$ converges, we have that $(s_{n})_{n=1}^{\infty}$&lt;br /&gt; converges.&lt;br /&gt; &lt;br /&gt; We use the ratio test. Let $b_{k}=a_{k}/s_{k}.$ Then $\lim_{k\rightarrow\infty}b_{k+1}/b_{k}=\lim_{k\rightarrow\infty}\dfrac{\dfrac{a_{k+1}}{s_{k+1}}}{\dfrac{a_{k}}{s_{k}}}=\lim_{k\rightarrow\infty}\dfrac{a_{k+1}s_{k}}{a_{k}s_{k+1}}=\lim_{k\rightarrow\infty}\dfrac{a_{k+1}}{a_{k}}\cdot\lim_{k\rightarrow\infty}\dfrac{s_{k}}{s_{k+1}}$&lt;br /&gt; &lt;br /&gt; By the ratio test, since $\sum_{k=1}^{\infty}a_{k}$ converges, $\lim_{k\rightarrow\infty}\dfrac{a_{k+1}}{a_{k}}=L,$&lt;br /&gt; for some $L&amp;lt;1.$&lt;br /&gt; &lt;br /&gt; Also, since $(s_{n})_{n=1}^{\infty}$ converges, we have that $\lim_{k\rightarrow\infty}\dfrac{s_{k}}{s_{k+1}}=1.$ &lt;br /&gt; &lt;br /&gt; Then $\lim_{k\rightarrow\infty}\dfrac{a_{k+1}}{a_{k}}\cdot\lim_{k\rightarrow\infty}\dfrac{s_{k}}{s_{k+1}}=M$&lt;br /&gt; for some $M&amp;lt;1,$ so by the ratio test, $\sum_{k=1}^{\infty}a_{k}/s_{k}$&lt;br /&gt; converges.<br /> <br /> possible problems with my proof (of which there are at least a few that i am aware of)<br /> 1: ratio test is not an &quot;if and only if&quot;, so i don&#039;t know i can make the conclusion i did that lim a_k+1/a_k &lt; 1.<br /> 2: i never really used the fact that a_k &gt; 0 for all k. I&#039;m tempted to say that because of that fact, s_k/s_k+1 &lt; 1 for all k, so multiplying that by lim a_k+1/a_k, which is also less than 1 results in a number &lt; 1.<br /> 3: i didn&#039;t really show that lim s_k/s_k+1 = 1. it seems intuitive to me that it is, but without having anything actually defined, I&#039;m sort of at a loss to actually prove it.<br /> <br /> any guidance would be greatly appreciated.

 

[bennyska]

typo in the pdf. nothing serious. sum should be to n.

 

[eczeno]

benn is correct. when you define s_n, the sum should go to n, not infinity.

as for the proof itself, it is almost there, but you are right to be skeptical of your use of the ratio test. you have not accounted for the possibility that L=1 (which turns out to be pretty important for the rest of your proof).

as for the limit of \frac{s_k}{s_{k+1}}, it must be equal to 1, but we need to show it. this is not hard. since the limit of s_k and s_{k+1} must be equal if they exist, and they do exist.

 

[bennyska]

for some reason, my updated proof isn't showing up right (the new code is just showing up as the old problem), but i did a new proof using cauchy criteria. it's attached (the second proof is a different problem, but any critiquing on that would be appreciated too.

 

[eczeno]

just taking a quick look, i see one mistake in the new proof. you say that

a_k&lt;s_k for k&gt;1 implies that \frac{a_k}{s_k} &lt; a_k for k&gt;1

but this does not follow. note: s_k could be less than one.

 

[bennyska]

duh. should have seen that. i think i fixed it.

 

[eczeno]

hi benn,

i am with you until you write

\sum_{k=m+1}^n{a_k}&lt;B\epsilon

i don't know how the B found it's way in there.

and after that i get very confused, i am sorry to say. i am not quite sure what you mean by divide through by s_k. but, whatever you meant, what you wrote is not necessarily true.

 

[HallsofIvy]

1) When you "update" something, do NOT just edit the original post. People who have already looked at that post may not look at it again. Put your update in a new post in the thread.

2) The board has an annoying but minor problem with LaTeX- when you edit, it does not come out right. You may see exactly what you had before- but I have known it to mix up LaTeX from different parts of the post. Fortunately there is a simple fix- click on the "refresh" button on hour internet reader. I have gotten to where I do that automatically when I edit LaTeX without even reading it first.

 

[bennyska]

so I'm trying to show that the series a_k/s_k converges. so if can show cauchy criteria, then it works. since the cauchy criteria is true for a_k for all epsilon>0, then it should be true for b*epsilon, where b is a positive constant. i put the b in there so i could show cauchy holds for the series when i divide a_k by s_k. s_k is strictly increasing, so s_k < s_k+1. b = s_1. since s_k is increasing, a_k/s_k < a_k/b when k>1. maybe i should switch other order and say let B=s_1.
oh, i see i didn't correct the second part...

also, I'm not sure why my code is coming out so poorly. i type this stuff up in a latex editor and try to post the code in here, but it seems to not work out so well.
it's also attached, and more more readable.

Assume that $\sum\limits _{k=1}^{\infty}a_{k}$ converges with each<br /> $a_{k} &gt; 0.$ Let $s_{n}=\sum\limits _{k=1}^{n}a_{k}.$ Show that $\sum\limits _{k=1}^{\infty}a_{k}/s_{k}$<br /> converges.<br /> <br /> Since $\sum\limits _{k=1}^{\infty}a_{k}$ converges, it satisfies<br /> the Cauchy criteria. Also, since $a_{k} &gt; 0 for all $k\in\mathbb{R},$<br /> we have that $|a_{k}|=a_{k},$ and $s_{k}&gt;a_{k}$ for all $k&gt;1.$<br /> Let $B=s_{1},$ where $B&gt;0.$ Since $a_{k}&gt;0$ for all $k,$ $(s_{k})_{k=1}^{\infty}$<br /> is a strictly increasing sequence.<br /> <br /> By Cauchy criteria, for all $\varepsilon&gt;0,$ there exists an $n_{0}\in\mathbb{N}$<br /> such that if $n&gt;m\geq n_{0},$ $|\sum\limits _{k=m+1}^{n}a_{k}|=\sum\limits _{k=m+1}^{n}a_{k}&lt;B\varepsilon.$ <br /> <br /> So $a_{m+1}+a_{m+2}+...+a_{n}&lt;B\varepsilon.$ Dividing through by<br /> $s_{k},$ we have $\sum\limits _{k=m+1}^{n}\dfrac{a_{k}}{s_{k}}=$$\dfrac{a_{m+1}}{s_{m+1}}+\dfrac{a_{m+2}}{s_{m+2}}+...+\dfrac{a_{n}}{s_{n}}&lt;\dfrac{a_{m+1}}{B}+\dfrac{a_{m+2}}{B}+...+\dfrac{a_{n}}{B}=\dfrac{1}{B}(a_{m+1}+a_{m+2}+...+a_{n})&lt;\dfrac{B\varepsilon}{B}=\varepsilon,$<br /> so $\sum\limits _{k=1}^{\infty}a_{k}/s_{k}$ converges by Cauchy criteria

 

[eczeno]

ok, i see where you are going now, but the B is still a problem for me, though i think it is an easy fix. if you use epsilon in the first part of the theorem, you must use epsilon in the conclusion as well, especially since B could be very small. i would just define delta = B*epsilon and use delta for the sum on a, then you will be left with epsilon to use when you apply cauchy to prove the sum on a/s. i think that might do it.