Question

\[\sum_{n=1}^{\infty} \frac{n^{2}+1}{n^{3} +1}\]
my book did a limit comparison test for this
\[\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}}\]
and somehow they took an n/1 and multiplied it to
\[\frac{n^{2}+1}{n^{3} +1}\]
and made this sum divergent, i believe its supposed to be convergent

Answer 1

\[ \frac{n^{2}+1}{n^{3} +1} = \frac{n^{2}}{n^{3} +1} + \frac{1}{n^{3} +1} \approx \frac{1}{n} + \frac{1}{n^{3} +1}\]

1/n diverges, and 1/n^3+1 converges.

Answer 2

how does
\[\frac{n^{2}}{n^{3} +1} \approx \frac{1}{n} \]

Answer 3

i guess dividing numerator and denominator by 1/n^2

Answer 4

as n tends to infinity,
\[ \lim_{n \rightarrow \infty } \frac{1}{n + \frac 1 {n^2}}\]
Yup ..

Answer 5

if you want you can perform integral test .. i don't think ratio test will be conclusive. also try pluggin in wolf

Answer 6

ok so 1/n is divergent ...why? \[1/\infty\] should be convergent

Answer 7

why do you think ... ?? ratio test in inconclusive ... from integral test you find 1/n is divergent. You can prove it shortly if you want. like -1+1-1+1-1+1 ..

Answer 8

ok i'll use the integral test

Answer 9

Let 1/n converge to x say
1+1/2+1/3+ 1/4 + 1/5 +1/6 + ... = x
1/2 + 1/4 + 1/6 + .... 1 + 1/3 +1/5 + 1/7 + ... = x
1/2(x) + 1 + 1/3 +1/5 + 1/7 + ... = x
or, 1 + 1/3 +1/5 + 1/7 + ... = 1/2 (x) = 1/2 + 1/4 + 1/6 + ...
which is a contradiction ... so 1/n is divergent.

Answer 10

i see...back to the limit comparison test for a second...what about this
\[\lim_{n\rightarrow\infty} \frac{a_{n}}{b_{n}}\] ordeal

Answer 11

where \[a_{n} =\frac{n^{2}}{n^{3} +1}\]
and \[b_{n}= 1/n\]

Answer 12

sorry an =\[\frac{n^{2}+1}{n^{3} +1} \]

Answer 13

take n tends to infinity ... try pluggin in wolframalpha ... but i don't think ratio test will be conclusive.