anonymous
  • anonymous
Does the series converge or diverge? Sum 10^((-n^2)/(n+1)), n-> 1 to infinity. I "think" its the root test, but kinda stuck.
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
I went ahead and looked at the exponent and factored out an n so the series becomes 10 to the -(n/(1+(1/n))) if I take the limit of the exponent as it goes to infinity it is cear that it goes infinitly high. Now because it is negative this makes the sum something like 1/(10^(infinity)) which would be going to the number zero. I would say it converges to zero.
anonymous
  • anonymous
if it kinda root test.. Let L = lim (n -- > inf) | an |1/n. If L < 1, then the series sum (1..inf) an converges. If L > 1, then the series sum (1..inf) an diverges. If L = 1, then the test in inconclusive
anonymous
  • anonymous
and actually that would be if the sequence convereged and what not, I think the fact that the limit exist with respect to the sum of the series suggests that it diverges. There are so many rules and what not with sequences and sums. Check the literature you have on it for clarity of the result, but it can be resolved by taking the limit. The approach is the same, but the matter of whether or not it diverges is the issue. The real answer I believe it is is that it is divergent b/c we are analyzing the sum and not the sequence.

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anonymous
  • anonymous
I believe daomowon that your first reply is the correct answer. I plugged it into wolfram alpha, and it says it does converge(wont say how of course). So if the L = 0 like you said, then it would converge. Regardless, thanks for the assist in showing me.
anonymous
  • anonymous
I don't know what is up with the site today, too many replies, not posted by choice. It is divergent though. I just spoke too fast on the initial approach.
anonymous
  • anonymous
whoa, wait my bad, I almost forgot that it was less than 1 and all that, I guess becaue I was back on that whole limit existing thing. Sometimes if a limit does exist it makes the sum diverge, You're right abuot it Net. My bad.
anonymous
  • anonymous
Sequences and sums are weary just because one rule doesn't necessarily dictate the whol deal, and things change from sequnce to series, and all that too. Oh well.
anonymous
  • anonymous
Aye, this one was being a pain for me. Still, thanks again.

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