Question

Which series convergence cannot be determined by direct comparison with a geometric series or p-series?

Answer 1

\[\sum_{n=1}^{\infty} \frac{ 1 }{ n2^n } \]
\[\sum_{k=1}^{\infty} \frac{ k }{ k!+4^k } \]
\[\sum_{n=1}^{\infty} \frac{ \sin^2n }{ n^2 } \]
\[\sum_{k=1}^{\infty}\frac{ k ^{1/3} }{ k^2+k} \]
\[\sum_{n=3}^{\infty} \frac{ 3n+5 }{ n(n-1)(n-2) }\]
\[\sum_{n=2}^{\infty} \frac{ n^2 }{ n^4 -1 }\]

Answer 2

how do i do this?

Answer 3

one at a time

Answer 4

\[\sum_{n=1}^{\infty} \frac{ 1 }{ n2^n }\] each term is less than \(\frac{1}{2^n}\) a convergent geometric series right?

Answer 5

\[\sum_{n=1}^{\infty} \frac{ \sin^2n }{ n^2 }\] compare with \(-\frac{1}{n^2}\) and \(\frac{1}{n^2}\) then squeeze

Answer 6

oh nvm numerator is positive, compare only with \(\frac{1}{n^2}\)

Answer 7

i have no idea how you do this one\[\sum_{k=1}^{\infty} \frac{ k }{ k!+4^k }\] although convergence is pretty obvious

Answer 8

why is that obvious?

Answer 9

because both \(k!\) and \(4^k\) grow way way way (way) faster than \(k\)
so together they are like overkill

Answer 10

problem you asked though is to compare to p series or geometric, series, and i don't think they compare to either

Answer 11

we can check the others if you like, but they all look like you could compare to p series to me

Answer 12

how would i tell if the convergence cannot be determined?

Answer 13

like for example, what you would compare this to \[\sum_{k=1}^{\infty}\frac{ k ^{1/3} }{ k^2+k}\]

Answer 14

they are not asking if convergence cannot be determined
they are asking if it cannot be determines by comparison to p series or geometric series

Answer 15

so is it clear what you would compare \[\sum_{k=1}^{\infty}\frac{ k ^{1/3} }{ k^2+k}\] to easily?

Answer 16

k/k^2 ? :/

Answer 17

hmm no

Answer 18

first off \(\frac{k}{k^2}=\frac{1}{k}\)
maybe it is not clear but we can make it so
the degree of the numerator is \(\frac{1}{3}\) and the degree of the denomnator is \(2\) so you would compare it to \[\large\frac{k^{\frac{1}{3}}}{k^2}=\frac{1}{k^{2-\frac{1}{3}}}=\frac{1}{k^{\frac{5}{3}}}\]

Answer 19

which is a p series with \(p>1\) so it converges
`

Answer 20

similarly this one \[\sum_{n=2}^{\infty} \frac{ n^2 }{ n^4 -1 }\] compare with \[\frac{n^2}{n^4}=\frac{1}{n^2}\]