Math Contest #20 [2 SBI] <<<10 STEM for you if you solve the bonus problem>>>

in #puzzle4 years ago

Here you can keep your brain fit by solving math related problems and also earn SBI or sometimes other rewards by doing so.
The problems usually contain a mathematical equation that in my opinion is fun to solve or has an interesting solution.
I will also only choose problems that can be solved without additional tools(at least not if you can calculate basic stuff in your head), so don't grab your calculator, you won't need it.
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Rules

No upvote, No resteem, No follow required!

I will give the SBI(s) randomly to any participants.

I will give 10 STEM to every one who gets the bonus question.

You have 4 days to solve it.

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Problem

After the infinite sum last week I want you to do a sequence today:
Screenshot from 2019-10-26 22-11-59.png

  1. Does the sequence converge?
  2. If yes what is the limit of the sequence?
  3. Bonus(10 STEM if you solve it!): Can you construct a similar sequence that converges to √n, where n is your reputation?

To 3.:
I ask for your the square root of your reputation here because I want to give everyone the opportunity to get a result by thinking for themselves and not copying from others. So please just enter the result for 3 without any extra comment.

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To everyone who already participated in a past contest, come back today and try a new problem(tell me if you don't want to be tagged):
@addax @ajayyy @athunderstruck @bwar @contrabourdon @crokkon @fullcoverbetting @golddeck @heraclio @hokkaido @iampolite @masoom @mmunited @mobi72 @mytechtrail @ninahaskin @onecent @rxhector @sidekickmatt @sparkesy43 @syalla @tonimontana @vote-transfer @zuerich

In case no one gets a result(which I doubt), I will give away the prize to anyone who comments.

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@contrabourdon sponsors my contests with 2 STEEM weekly.
You can support him by using a witness vote on untersatz, so he can further support this and other contests.

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  1. & 2. Using simple estimations one can show that the series is limited to [1,2]. 2/a_n has the property that for a_n>sqrt(2) the term will be <sqrt(2) and the other way round. Since you effectively get the mean between a_n and 2/a_n each iteration, this will bring you closer and closer to sqrt(2).
  2. Example: every number b>1 will definetly work and the series will converge to sqrt(b). a(n+1)=0.5(a(n) + b/a(n)) will give b as converging value.

This post has been upvoted by witness @untersatz. You've done a great job!
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