Question

Series question

Answer 1

\[\sum_{n=2}^{\infty} \frac{ 1 }{ (\ln n)^{\ln n} }\]

Answer 2

series of serious... question..?

Answer 3

\[a_n = \frac{ 1 }{ (\ln n)^{\ln n} }\]

\[b_n = \frac{ 1 }{ n^2 }\] ( I just picked this, but I'm not exactly sure for comparison tests how to pick a bn would be nice if someone could tell me)

\[\frac{ 1 }{ n^2 } > \frac{ 1 }{ (\ln n)^{\ln n} }\]

\[(\ln n )^{\ln n} > n^2 \implies \ln (\ln n)^{\ln n} > \ln n^2\]

\[\ln n \ln(\ln n) > 2 \ln n \implies \ln(\ln n) > 2 \implies e^{\ln(lnn)} > e^2 \implies \ln n > e^2 \implies e^{\ln n} > e^{e^{2}}\]

Answer 4

\[e^{lnn} > e^{e^{2}} \implies n > e^{e^{2}}\] as it got cut off

Answer 5

Not sure at what step I take the limit haha...

Answer 6

After that could I just say lim n-> infinity ln(n) -> infinity and \[\sum_{n=2}^{\infty} \frac{ 1 }{ n^2 } \] p=2>1 converges so the whole series converges?

Answer 7

@Kainui

Answer 8

@ganeshie8