Question

Given that SUM_{n=1}^{infinity}(u_n) is a divergent series of positive terms, show that SUM_{n=1}^{infinity}(u_n)^p diverges for 0 < p < 1.

Answer 1

That is, \[\sum_{n=1}^{\infty}u_n\] diverges is given. I need to show that \[\sum_{n=1}^{\infty}u_n^p\] also diverges for 0<p<1.

Answer 2

"comparison test" i believe is the term of art

Answer 3

since
what jamesj said.

Answer 4

\[u_n\] are positive, and \[u_n^p > u_n\] is only true if \[u_n < 1\] which isn't something one can assume...

Answer 5

yes sorry. It's more subtle than that, one sec.

Answer 6

hold on. if after some n
\[u_n>1\] it diverges for sure right?

Answer 7

in which case you are done

Answer 8

yeah, but we are not given that the \[{u_n}\] form an increasing sequence...

Answer 9

Right, thank-you. If it's not true that there is some N such that
\[ n > N \implies |u_n| < 1 \]
then u_n^p stays greater than 1 also and hence diverges.

But if it is true that the tail of (u_n) goes under 1, then you can use the fact that
\[ u_n^p > u_n \]
for those terms.

Answer 10

Maybe they do as a consequence of the given diverging sum?

Answer 11

Yeah James I think that you are right...

Answer 12

only the tail determines convergence or divergence

Answer 13

In other words:
\[ u_n \geq 1 \implies u_n^p \geq 1 \]
\[ u_n < 1 \implies u_n^p > u_n \]

hence we may bound every term of \( u_n^p \) by
\[ u_n^p \geq \min(1,u_n) \]
and the sum of those terms has to diverge.

Answer 14

it does say the are positive terms. if there does not exist and N for which all n>N you have
\[u_n<1\] then it diverges because the terms do not even go to zero. so we can assume then that there does exist such an N and use what jamesj said above, i.e. comparison test

Answer 15

more eloquently put by jamesj

Answer 16

Very good, thanks both of you!

Answer 17

and I promise to drink at least one cup of coffee before attempting such questions in the future.