Question

tried ratio test got messy
gave up
halp

Answer 1

Actually ratio should work

Answer 2

\[a_n = \left( \frac{ n^2+1 }{ 2n^2+1 } \right)^n\]
\[a_n+1 = \left( \frac{ (n+1)^2+1 }{ 2(n+1)^2+1 } \right)^{(n+1)}\] \[\lim_{n \rightarrow \infty} \left| \frac{ a_n+1 }{ a_n } \right|\] it should converge

Answer 3

\[a_{n+1}\]

Answer 4

another way you might be interested in
compare it with this series
\(\Large \sum_{n=1}^{\infty}(\frac{1}{2})^n\)

Answer 5

Nice,
\[n^2+1\lt n^2+2 \\~\\\implies n^2+1\lt \dfrac{1}{2}(2n^2+1) \\~\\\implies \dfrac{n^2+1}{2n^2+1}\lt \dfrac{1}{2}\]

Answer 6

If that doesn't work, consider (2/3)^n...
comparison test is all about guessing the magic series that works just right :)

Answer 7

for \(n\ge 2\) we have :

\[1 \lt \dfrac{n^2}{3} \\~\\\implies n^2+1\lt n^2+\dfrac{n^2}{3} \lt \dfrac{4}{3}n^2\lt \dfrac{4}{3}n^2+\dfrac{2}{3} =\dfrac{2}{3}(2n^2+1)\\~\\\implies \dfrac{n^2+1}{2n^2+1}\lt \dfrac{2}{3}\]

Answer 8

It follows \[\left(\dfrac{n^2+1}{2n^2+1}\right)^n\lt \left(\dfrac{2}{3}\right)^n\]

Answer 9

lol any geometric series with base r >1/2 works :P

Answer 10

Yes

Answer 11

i'm not asking... i'm js