osanseviero
  • osanseviero
Determine if the next progression have a limit (demostrate it) and determine it
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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osanseviero
  • osanseviero
\[\left( -1^{n} \right)\left( \frac{ 5n+4 }{ 2n } \right)\]
osanseviero
  • osanseviero
Help, please?
osanseviero
  • osanseviero
:/

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tkhunny
  • tkhunny
As n increases in the positive direction, \(\dfrac{5n+4}{2n}\) approaches \(\dfrac{5}{2}\). Follow with your mind as n increases. The little 4 on the end of the numerator becomes less and less significant. The terms do NOT approach zero.
osanseviero
  • osanseviero
oh...I think I see it now...but how can i demostrate if it is it's limit?
osanseviero
  • osanseviero
but there is also the -1^n
osanseviero
  • osanseviero
So it approaches 5/2 and -5/2 ?
tkhunny
  • tkhunny
The typical demonstration is a division by n. For n > 0, \(\dfrac{5n+4}{2n} = \dfrac{5 + \dfrac{4}{n}}{2}\). In this form, it is relatively obvious that the limit it 5/2 as n increases. The FIRST criterion for convergence is terms that approach ZERO. Nothing else will do. These terms do not approach zero, therefore, we do not care about the alternating sign. If the terms approach zero, THEN we'll worry about the sign.
osanseviero
  • osanseviero
what I mean is that there is a (-1^n) multiplying all of that...so -5/2 is also a limit
tkhunny
  • tkhunny
No, this is not a limit. Limits come alone, not in pairs. The terms, without the sign, approach 5/2. I may have stated that carelessly, before. The actual terms, including the sign, do not have a limit. They is oscillating.
osanseviero
  • osanseviero
Oh...okk
osanseviero
  • osanseviero
So for this there isnt a limit, neither \[\frac{ 1 }{ 2 },2^{2}, \frac{ 1 }{ 2^{3} }\]
tkhunny
  • tkhunny
Seriously? An alternating sign in the exponent? No. No limit.
osanseviero
  • osanseviero
I thought so, thanks :)

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