Series 6-3/2-3/4=3/8-...., write general term a_n, determine

[TheTruth319]

Hey guys im new to this so im trying to get a feel of how this work. Were working on sequences and series and im having major trouble with some problems.

1. Consider the series: 6-3/2-3/4=3/8-....

a. write down the general term a_n of the series
b. Determine if the series is convergent and if yes find its limit.

 

[mmm4444bot]

Re: Series

... im trying to get a feel of how this work.

Hello Truth:

I'm guessing that you've used the pronoun "this" to mean "this web site".

Did you read the post titled, "Read Before Posting"? It outlines your responsibilities for seeking help at this site.

Please tells us what you already know about solving exercises like this one. Also, show any work that you've been able to accomplish. (We do not, in general, provide lessons at this site.)

Do you have any specific questions about how to finish this exercise?

I would start by factoring out the negative threes. That yields an easily-recognizable geometric series.

Do you know how to sum an infinite series?

Please, give us a clue as to what you're thinking.

Cheers,

~ Mark

 

[stapel]

1. Consider the series: 6-3/2-3/4=3/8-....

Um... This doesn't look like a series; it's more of an unfinished addition (I think): 6 - 3/2 - 3/4 = 24/4 - 6/4 - 3/4 = (24 - 9)/4 = 15/4.

Please reply with clarification, showing all of your work and reasoning so far. Thank you!

Eliz.

 

[soroban]

Hello, TheTruth319!

Did you ever look at what you typed?

One little typo and you've lost everyone . . .

\(\displaystyle \text{1) Consider the series: }\:6-\tfrac{3}{2} - \tfrac{3}{4} - \tfrac{3}{8} - \hdots\)

a. Write down the general term \(\displaystyle a_n\) of the series

b. Determine if the series is convergent and if yes, find its limit.

\(\displaystyle \text{The series seems to be: }\:S \;=\;6 - \tfrac{3}{2} - \tfrac{3}{2^2} - \tfrac{3}{2^3} - \hdots\)

\(\displaystyle \text{(a) The general term is: }\:\boxed{a_n \:=\:6 - \frac{3}{2^{n-1}}}\)



\(\displaystyle \text{We have: }\:S \;=\;6 - \tfrac{3}{2}\underbrace{\left(1 + \tfrac{1}{2} + \tfrac{1}{2^2} + \tfrac{1}{2^3} + \hdots \right)}_{\text{geometric series}}\)

\(\displaystyle \text{The geometric series has: }\:\text{first term }a = 1\text{, and common ratio }r = \tfrac{1}{2}\)
. . \(\displaystyle \text{Its sum is: }\:\frac{1}{1-\frac{1}{2}} \:=\:2\)

\(\displaystyle \text{Therefore: }\;S \:=\:6 - \tfrac{3}{2}(2) \;=\;\boxed{3}\)

 

[mmm4444bot]

Re:

...This doesn't look like a series ...

I agree, but The Truth told us that it is a series. Do you believe The Truth? :lol:

6 - 3(1/2 + 1/4 + 1/8 + 1/16 + ...)

6 - 3(1) = 3

(b) The series is convergent, and its limit is 3.

Does The Truth understand the definition of the term "series"? (This is a rhetorical question.)

~ Mark