Which sum is larger?

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Randyyy

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\[ \sum_{k=0}^{100} 3^{-k}Cos(\frac{k \pi}{2}) \], \[ \sum_{k=0}^{200} 3^{-k}Sin(\frac{k \pi}{2}) \]
I have to prove that one of these sums are about 3 times as larger as the other.

Using the unit circle I figured out this pattern for Cosine, (1,0,-1,0) and for Sin(0,1,0,-1)
this means we can ignore 50 summations for cos and 100 for sin since they will be 0.

For cosine I wrote out the first 4 summations, [MATH]1+0-3^{-2}+0+3^{-4}...[/MATH] Removing the 0s and moving the 1 to the front as well as realizing we will get the same summation of [MATH]-3^{-2}+3^{-4}[/MATH] every time except it will be [MATH]\frac {1}{3^2}[/MATH] Smaller, I rewrite the sum as:
\[ 1+\sum_{k=0}^{50} (\frac{-3^{-2}+3^{-4}}{3^{2k}})\]
I then do the same for Sin but It doesn´t feel like I´ve proven anything. Sure after 2-3 summations I get the correct answers but is this really what was intended? That I simplify or rewrite the summations and take the sum of 2-3 iterations and then compare? From this I get the cos sum to be about 0.9 and the Sine one to 0.3 exact.

Sin was rewritten to \[ \sum_{k=0}^{100} (\frac{3^{-1}-3^{-3}}{3^{4k}})\] Using the same logic.

Is my method correct or did I completely miss the "proving" part of the question? As I said, feels like I just rewrote it and although it shows me if i take 2-3 sums of each what the answer will be, I don´t know how I can argue that it is the finite value the sums will get.
 
Can you write these as (differences of) geometric series, and apply the appropriate formula to get exact values?
 
I´m guessing you want me to rewrite both series as this: [MATH]S_{k}= a \frac {r^{k}-1}{r-1}[/MATH] and then subtract them?
 
Randyyy said:
I´m guessing you want me to rewrite both series as this: [MATH]S_{k}= a \frac {r^{k}-1}{r-1}[/MATH] and then subtract them?
Click to expand...

Yes ... or, if you take my question literally, tell me that you can't.

I haven't checked all the details to make sure you didn't get a number wrong, but it you're right so far, then this seems like the next step if you want an actual proof.
 
Randyyy said:
I´m guessing you want me to rewrite both series as this: [MATH]S_{k}= a \frac {r^{k}-1}{r-1}[/MATH] and then subtract them?
Click to expand...
The formula is [MATH]S_{k}= a \frac {r^{k+1}-1}{r-1}[/MATH] or [MATH]S_{k-1}= a \frac {r^{k}-1}{r-1}[/MATH]
 
Dr.Peterson said:
Yes ... or, if you take my question literally, tell me that you can't.

I haven't checked all the details to make sure you didn't get a number wrong, but it you're right so far, then this seems like the next step if you want an actual proof.
Click to expand...

In all honesty I am not that good with geometric series, from what I know r should be the quotient of for example [MATH]\frac{a_{2}}{a_{1}}[/MATH] if the numerator is the 2nd value and denominator is the 1st value of the sum. Using this yields about 1.5 for the cos series when k= 5, which isn´t correct.
I did compare the values of the sums and it did show that the sine gave values 3 times as large but it´s hard to prove anything since I can´t really know what the sums are(I know only because I computed them prior).
for example, [MATH]-3^{-2}+3^{-4}= \frac {-8}{81}[/MATH] if i take sine now as well for k=0 it gives : [MATH]3^{-1}-3^{-3}=\frac{8}{27}[/MATH]and this was true for every value of k that the next pair of numbers sine had 3 times larger. the only issue is that cosine gives negative numbers. Although it is 1-......=0.9, it probably is not sufficient a a proof.

Jomo said:
The formula is [MATH]S_{k}= a \frac {r^{k+1}-1}{r-1}[/MATH] or [MATH]S_{k-1}= a \frac {r^{k}-1}{r-1}[/MATH]
Click to expand...
Oh, I tried that instead and it gave 1.8, It is clear that my r has got the wrong value, not too sure how I find what r is.


I got r to be [MATH]\frac{1}{9}[/MATH]
 
Maybe I am missing something but the numerator is not always 3^-2 - 3^-4

I get one sum to be 1/3^0 - 1/3^2 + 1/3^4 + ... + 1/3^100 (why is that last term positive??) = 1/9^0 - 1/9^1 + 1/9^2 - ... + 1/9^50 = what using the fact that this is a geometric series.

The other sum is 1/3^1 - 1/3^3 + 1/3^5 - ... 1/3^199 (why is the last term negative??) (1/3)(1/3^0 - 1/3^2 + 1/3^4 + ... + 1/3^198) = (1/3)(1/9^0 - 1/9^1 + 1/9^2 - ... + 1/9^99). This being a geometric series you can easily find this sum.

Please post back.
 
I am not sure why Dr Peterson wants you to subtract the results. Either he is seeing something that I am not seeing (not unusual) or he meant to say that you should divide the two results. After all, if on number is three times another number then when you divide the larger by the smaller you get 3.
 
Jomo said:
I am not sure why Dr Peterson wants you to subtract the results. Either he is seeing something that I am not seeing (not unusual) or he meant to say that you should divide the two results. After all, if on number is three times another number then when you divide the larger by the smaller you get 3.
Click to expand...

I mentioned differences only because he showed the numerators as differences, and could (if he wanted to waste time) break the sum into two sums. I didn't mean to subtract the cosine and sine series.

I think he has combined pairs of terms from the original series to avoid an alternating series, but as you show, that is unnecessary. I was going to start from his work, but it's too easy to get tangled up in it, so I abandoned an answer I was writing.
 
Okay, I think maybe rewriting the sums wasn´t too helpful like I originally did. So going back to the original sums (given in the question)
I need to find the sum using [MATH]S_{k-1}= a \frac {r^{k}-1}{r-1} [/MATH] or [MATH]S_{k}= a \frac {r^{k+1}-1}{r-1} [/MATH] if I understood correctly and then divide them which should yield 3 or 1/3 (depending on which I put as the numerator and denominator) if it is true that one is 3 times as big.
 
The problem never stated that one sum is 3 times the other sum. Please read it more carefully.
 
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