What is the convergence status of the given sequence and series?

  • Thread starter Thread starter ILoveBaseball
  • Start date Start date
  • Tags Tags
    Sequence Series
AI Thread Summary
The sequence defined by An = 5n/(12n+5) converges to 5/12 as n approaches infinity. For the series from n=1 to infinity of An, it diverges because the terms do not approach zero; instead, they approach 5/12. The absence of a common ratio indicates that the series cannot converge. Therefore, the series is classified as divergent. In conclusion, the sequence converges while the series diverges.
ILoveBaseball
Messages
30
Reaction score
0
An = \frac{5n}{12n+5}

For both of the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise.


a.) The series n=1 to infinity(i don't know how to make the 'sum of' sign)
(An) I'm having trouble with this one, there seems to be no 'common ratio', so does this mean it's divergent? A_1 = 5/17, A_2 = 10/29, A_3 = 15/41

b.) the squence A_n. well for this one as n-> infinity, the limit should be 5/12
 
Physics news on Phys.org
ILoveBaseball said:
An = \frac{5n}{12n+5}

For both of the following answer blanks, decide whether the given sequence or series is convergent or divergent. If convergent, enter the limit (for a sequence) or the sum (for a series). If divergent, enter INF if it diverges to infinity, MINF if it diverges to minus infinity, or DIV otherwise.


a.) The series n=1 to infinity(i don't know how to make the 'sum of' sign)
(An) I'm having trouble with this one, there seems to be no 'common ratio', so does this mean it's divergent? A_1 = 5/17, A_2 = 10/29, A_3 = 15/41

b.) the squence A_n. well for this one as n-> infinity, the limit should be 5/12


Your b) looks good and should tell you the answer to a) What is 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 + 5/12 +. . .
 
If

A_n = \frac{5n}{12n+5}

then

\lim_{n \rightarrow \infty} A_n = \lim_{n \rightarrow \infty} \frac{5}{12+5/n}
=\frac{5}{12+(\lim_{n \rightarrow \infty} 5/n)} = 5/12
 
Hint for a) In order for a series such as \Sigma_1^{\infinity}A_n to converge, it is necessary that the sequence {An} converge to 0.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top