Question

calculus- does this converge, diverge, or is inconclusive by ratio or limit comparison test?

Answer 1

\[\lim_{n \rightarrow \infty} \frac{ a _{n} }{ n} = 0.8\]

Answer 2

\[\lim_{n \rightarrow \infty} n ^{2}a _{n}=8\]

Answer 3

what is this exactly?

Answer 4

is "this" \[\sum_{n=1}^{\infty} a_n \]?

Answer 5

yes

Answer 6

does it coverge/diverge by the rario test
or
converge/diverge by the limit comparison test
or we cannot say anything

Answer 7

I did this...
\[\lim_{n \rightarrow \infty}(n^2 a_n-\frac{a_n}{n})=8-.8=7.2 \\ \lim_{n \rightarrow \infty} a_n(n^2-\frac{1}{n})=7.2 \\ \lim_{n \rightarrow \infty} a_n(\frac{n^3-1}{n})=7.2 \\ \lim_{n \rightarrow \infty} \frac{n^3-1}{b_n n}=7.2 \text{ where } a_n=\frac{1}{b_n} \\ \\ \text{ I made the assumption } \\ \text{ that } b_n n \text{ was a polynomial } \\ \text{ and it would have to be a 3rd degree polynomial for the limit \to exist } \\ \text{ which is does } \\ b_n n=c n^3+d n^2+e n+f \\ \]
\[\text{ so we know } \frac{1}{c}=7.2 \implies c=\frac{1}{7.2}=\frac{10}{72}=\frac{5}{36} \\ \]

\[b _ n n=\frac{5}{36} n^3+dn^2+e n +f \\ b_n=\frac{5}{36} n^2+dn+e+\frac{f}{n} \\ a_n=\frac{1}{\frac{5}{36} n^2+dn+e+\frac{f}{n}} \\ a_n=\frac{36n }{5n^3+dn^2+en+f}\]

Then you should be able to use comparison test...

Answer 8

but I don't know if there were other possible assumptions

Answer 9

@ganeshie8 here is another one for you :p

Answer 10

these appear to be separate problems

Answer 11

also my sequence fails at least one of the limits

Answer 12

if \(\displaystyle\lim_{n \rightarrow \infty} \frac{ a _{n} }{ n} = 0.8\) then \(a_n\to\infty\) as \(n\to \infty\)

if \(\displaystyle\lim_{n \rightarrow \infty} n^2a _{n} =8\) then \(a_n\to 0\) as \(n\to \infty\)

Answer 13

I wonder if we can do something similar to what I did assume a_n is a first degree polynomial in the first

and assume a_n is a rational 1/(second degree polynomial) in the second

Answer 14

thanks for pointing out that it was 2 different questions
i didn't see that at first @Zarkon

Answer 15

for the first one we have that \[\sum_{n=0}^{\infty}a_n\] diverges by the divergence test (or we can use the comparison test)


for the second one the sumeconverges by the comparison test

Answer 16

*sum converges

Answer 17

\[\lim_{n\to\infty}\frac{a_n}{\frac{1}{n^2}}=\lim_{n\to\infty}n^2a_n=8\]

Answer 18

since the limit is a finite positive number we have that since \[\sum_{n=1}^{\infty}\frac{1}{n^2}\]
converges we have that \[\sum_{n=1}^{\infty}a_n\]

converges

Answer 19

why does the first one diverge?

Answer 20

\(a_n\to\infty\)


a necessary condition for a series to converge is that \(a_n\to 0\) as \(n\to\infty\)

Answer 21

ahh ok thanks