Sequence Converge/Diverge Problems

[timid summit]

Can anyone tell me which of these is wrong?

[ancient rose]

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[slender edge]

Is this the full problem? Also how many of the 7 are wrong?

[timid summit]

yes this is the full problem. i'm not sure which of the 7 are wrong, but i mainly need help on 3, 4, and 6.

[desert wren]

3 is right, 4-5-6 arent

for 4 you can maybe just plug infinity in for n
for 6, note that (n+1)/n has a limit of 1, so (n+1)^100/n^100 also has a limit of 1

so the numerator is basically multiplying by 1 as you go from term to term

[slender edge]

For 6 I’d use L’hopitals 100 times to get 100!/(ln(1.01)^100) • 1/(1.01)^n and show how 1.01^n approach infinity

[desert wren]

that also works

[timid summit]

im not really understanding 6. can you elaborate more on how to approach the problem?

[desert wren]

this is using ratio test, to prove that the limit of the ratios is less than 1;
doubleg is using lhop, because both the numerator and denominator go to infinity, so you can take the derivative 100 times

[timid summit]

ok i see so using the ratio test shows that the sequence converges to 1.

and for 4, by plugging in infinity that just shows that the sequence diverges to infinity?

[desert wren]

yes

[desert wren]

no, to 0

maybe you misunderstood what i was saying

the ratio between successive numerators converges to 1, the ratio between successive denominators is always 1.01

so the ratio between successive terms converges to 1/1.01

[timid summit]

ok i see thank you so much!

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