Question

Limit Comparison Test. See first post.

Answer 1

\[\sum_{n=1}^{\infty} \frac{ 1 }{ \sqrt{n ^{3}+1} }\]
Determine whether the series above converges or diverges using the Limit Comparison Test.

With the little I understand about this I have figured out that:
\[\sum \frac{ 1 }{\sqrt{n ^{3}} }\]
would be the similar term
all the examples I find check this using the test that has an r but I am unsure what the r is and how to find it

Answer 2

@terenzreignz convergence divergence for you

Answer 3

You want to know whether or not
\[\sum \frac{ 1 }{\sqrt{n ^{3}} }\]
is convergent?

Answer 4

That would be the first test I need to find and then that would be the bn over an

Answer 5

Don't bother.
Go for the integral test.
I'm sure you must be aching for more of that :P

Answer 6

or rather an/bn=L

Answer 7

Unfortunately this example specifies the need to answer using the Limit Comparison Test

Answer 8

The LCT is trivial for this one...that is what i would use

Answer 9

Of course. I just meant use the integral test (if you need confirmation) that
\[\Large \sum\frac1{\sqrt{n^3}}\] is convergent.
And after you convince yourself... use LCT with this and your original series.

Answer 10

Ah ok so for this segment:
\[\int\limits_{1}^{\infty}\frac{ 1 }{ \sqrt{n ^{3}} }\]
\[\frac{ -2 }{ \sqrt{n} } evaluated 1 \to \inf \]
\[\lim_{t \rightarrow \infty} (\frac{ -2 }{\sqrt{t} }+\frac{ 2 }{ \sqrt{1} })\]
\[0+2=2\]
so it is convergent

Answer 11

right, so you know that it's convergent, now try the LCT, IE, find this limit:
\[\Huge \lim_{n\rightarrow\infty} \left|\frac{\frac1{\sqrt{n^3+1}}}{\frac1{\sqrt{n^3}}}\right|\]

Answer 12

What are you using for the equations they are easier to read?

Anyway so:
\[\lim_{n \rightarrow \infty} \left| \frac{ \sqrt{n ^{3}} }{ \sqrt{n ^{3}+1} } \right|\]

if i dived top and bottom by n^3/2 i get 1/(1+0) when i evaluate so 1

Answer 13

\large \Large \LARGE \huge \Huge
^varying degrees of font size, type these before typing equations, to your liking.

Anyway, since the limit is 1 ( a positive number [which isn't infinity])
Then either both series diverge or both converge, but one of them, you know for a fact to be convergent, therefore...?

Answer 14

therefor they are convergent thanks much again might have another in a few mins