Finding the sum of an infinite series
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Steven G
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Yoh have to had tried something. Did you only learn about geometric series?
Write out a few terms and see if you can see something? Please try, you can get somewhere and someone here will push you a bit further.
If n was replaced with 1 what would the sum be?
Write out a few terms and see if you can see something? Please try, you can get somewhere and someone here will push you a bit further.
If n was replaced with 1 what would the sum be?
Dr.Peterson
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Evidently you have learned about geometric series, which this is not. In order to help you, we'd like to know what you have learned about series; please tell us, both specific types and techniques.
If you learned not only a formula for geometric series, but also how to derive that formula (or find the sum without a formula), then try using that same technique here, even though you don't know for sure that it will work.
This is an important way you learn math -- by being pushed a little beyond your knowledge, and forced to experiment rather than follow routines. Let us know what you find (in addition to what you have learned that we can use in helping).
In this case the sum would be 2.
I don't mean to be demanding help - I am just in the middle of studying for a test and feeling stuck
oh - figured it out (with some help from https://math.stackexchange.com/questions/1325254/what-does-sum-k-0-infty-frack2k-converge-to 
) It has to do with taking the derivative of the identity known from the geometric function and putting \(\frac12\) in as x.
) It has to do with taking the derivative of the identity known from the geometric function and putting \(\frac12\) in as x.
Dr.Peterson
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That's not the experiment I had in mind; "seems to" can lead toward an answer, but should not convince you.
oh - figured it out (with some help from https://math.stackexchange.com/questions/1325254/what-does-sum-k-0-infty-frack2k-converge-to
That's a different way to do it. If that's what you have been taught in your course, then definitely do it. Since you haven't told us what you learned, I can't tell.
My expectation was that you would multiply the summation by 2 and subtract, as we do with geometric series.
It's unfortunate that the internet gets in the way of thinking for oneself. I see far too many students who think that the way to solve a problem is to search for somebody telling them the answer, rather than searching for a way to figure it out themselves, using what they have been taught ...
It's true - but I think you are forgetting how helpful it is to ask someone for help after much time has been spent on searching for an answer....
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If I were going to solve this problem, I'd look at the first several sums:
[MATH]S_1=\sum_{k=1}^{1}\left(\frac{k}{2^k}\right)=\frac{1}{2}[/MATH]
[MATH]S_2=\sum_{k=1}^{2}\left(\frac{k}{2^k}\right)=\frac{1}{2}+\frac{2}{4}=\frac{2\cdot1+2}{4}[/MATH]
[MATH]S_3=\sum_{k=1}^{3}\left(\frac{k}{2^k}\right)=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}=\frac{4\cdot1+2\cdot2+3}{8}[/MATH]
Okay, it would appear we have:
[MATH]S_n=\frac{2^{n+1}-(n+2)}{2^n}[/MATH]
This will be our induction hypothesis \(P_n\). For our induction step, let's add [MATH]\frac{n+1}{2^{n+1}}[/MATH]...
[MATH]S_{n+1}=\frac{2^{n+1}-(n+2)}{2^n}+\frac{n+1}{2^{n+1}}=\frac{2\left(2^{n+1}-(n+2)\right)+n+1}{2^{n+1}}=\frac{2^{(n+1)+1}-((n+1)+2)}{2^{n+1}}[/MATH]
We have derived \(P_{n+1}\) from \(P_n\) thereby completing the proof by induction. Hence:
[MATH]S_{\infty}=\lim_{n\to\infty}S_n=2[/MATH]
[MATH]S_1=\sum_{k=1}^{1}\left(\frac{k}{2^k}\right)=\frac{1}{2}[/MATH]
[MATH]S_2=\sum_{k=1}^{2}\left(\frac{k}{2^k}\right)=\frac{1}{2}+\frac{2}{4}=\frac{2\cdot1+2}{4}[/MATH]
[MATH]S_3=\sum_{k=1}^{3}\left(\frac{k}{2^k}\right)=\frac{1}{2}+\frac{2}{4}+\frac{3}{8}=\frac{4\cdot1+2\cdot2+3}{8}[/MATH]
Okay, it would appear we have:
[MATH]S_n=\frac{2^{n+1}-(n+2)}{2^n}[/MATH]
This will be our induction hypothesis \(P_n\). For our induction step, let's add [MATH]\frac{n+1}{2^{n+1}}[/MATH]...
[MATH]S_{n+1}=\frac{2^{n+1}-(n+2)}{2^n}+\frac{n+1}{2^{n+1}}=\frac{2\left(2^{n+1}-(n+2)\right)+n+1}{2^{n+1}}=\frac{2^{(n+1)+1}-((n+1)+2)}{2^{n+1}}[/MATH]
We have derived \(P_{n+1}\) from \(P_n\) thereby completing the proof by induction. Hence:
[MATH]S_{\infty}=\lim_{n\to\infty}S_n=2[/MATH]
Dr.Peterson
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I didn't mean to criticize you specifically, but to lament that it is so easy either to just look for answers (as some do), or to search for help and find complete answers, shortcutting the process! When I was a student, there was no choice but to figure things out. (And, of course, asking someone for help is different from demanding ready-made answers.)
You are right
I accept the rebuke and can say that I'll try to ask for help not answers