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Mathematics 5 Online
OpenStudy (anonymous):

Find the exact sum of the following series

OpenStudy (anonymous):

\[\sum_{k=3}^{\infty} (1/2)^k\]

OpenStudy (anonymous):

Find the partial sum and take it's limit

OpenStudy (anonymous):

I think the answer is 1/4 since it's infinite geometric series

OpenStudy (anonymous):

how do you find the partial sum?

OpenStudy (anonymous):

ahh don't think that is required here unless you're taking a course in analysis

OpenStudy (anonymous):

so how do you know that is 1/4

OpenStudy (anonymous):

Google infinite geometric series; there's a formula for that

terenzreignz (terenzreignz):

Here... you know that geometric series take this form, right? \[\huge S=a + ar + ar^2+ar^3+...\] right? Where a is the first term and r is the common ratio.

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

Well, let's rearrange. \[\huge S-a=ar+ar^2+ar^3+ar^4+...\]following me so far? :)

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

Okay, let's divide both sides by r, the common ratio, we get... \[\huge \frac{S-a}{r}=a+ar+ar^2+ar^3+...\]Catch me so far?

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

Now, you'll notice that the right-side of this equation is nothing but S again. Right?

OpenStudy (anonymous):

yes

terenzreignz (terenzreignz):

So, we now have \[\huge \frac{S-a}{r}=S\]Now we just have to solve for S.

OpenStudy (anonymous):

a/(1-r)

terenzreignz (terenzreignz):

That's right.. \[\large S-a=Sr\]\[\large S-Sr=a\]\[\large S(1-r)=a\]\[\huge S=\frac{a}{1-r}\]

OpenStudy (anonymous):

so whats the "a" value for this series and whats the r value

terenzreignz (terenzreignz):

a is your first term r is what you keep multiplying to the first term, over and over again.

OpenStudy (anonymous):

so the first term is (1/2)?

terenzreignz (terenzreignz):

Your series starts at k=3, though.

OpenStudy (anonymous):

so (1/2)^3 is my first term?

OpenStudy (anonymous):

and r = (1/2)?

terenzreignz (terenzreignz):

Yep.

OpenStudy (anonymous):

thanks for your help

terenzreignz (terenzreignz):

No problem. :)

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