Question

Determine whether this infinite series diverges or converges...

Answer 1

\[\sum_{n=1}^{\infty}\frac{ n^3+1 }{ (n+1)^3 }\]

Answer 2

I believe it's converging if I remember my calc 2 correctly:
\[\sum_{n=1}^{\infty}\frac{ n+1 }{ (n+1)^2 } \approx \frac{ n^2 }{ n^2 }\approx 1\]

Answer 3

sorry i mistyped first part, supposed to be n^2+1, but that's my logic.

Answer 4

All exponents are 3, right?

Answer 5

Yeah sorry it's three, the same logic applies though. It will still simply to 1 with your approximation.

Answer 6

No, see, the question is whether the series is convergent, not whether the sequence is convergent, those are two different things...

Answer 7

Here, let me explain:
A sequence of 1's
\[\huge \left\{ 1 \right\}\]
is clearly convergent, this would approach 1, always, so this is convergent AS A SEQUENCE.

However
\[\huge \sum_{n=1}^{\infty}1\]
Does not converge at all, for what would you get if you add an infinite number of 1's ?

Answer 8

@terenzreignz Thanks for that, so that way to show what you have said,
\[\lim_{n \rightarrow \infty}\frac{ n^3+1 }{ (n+1)^3 }=1\]
So the series is divergent since the terms do not go to zero as n->inf.

Also, I am having problems with a similar question,
\[\sum_{n=1}^{\infty}\frac{ (n+1)! }{ (2n-1)! }\]

Answer 9

You're right about the first ; If a series is convergent, then it converges to zero AS A SEQUENCE (The converse is not true, ever heard of the Harmonic Series? :) )

Processing the second, hang on...

Answer 10

I have a hunch, but this situation seems to call for it; When factorial are involved, you might want to use the Ratio Test, heard of it?

Answer 11

Yep, I got it. Thanks for all your help @terenzreignz.
For anyone interested:
Apply the Ratio Test,
\[\lim_{n \rightarrow \infty}\left| \frac{ (n+2)! }{ (2n+1)! }*\frac{ (2n-1)! }{ (n+1)! } \right|\]
\[=\lim_{n \rightarrow \infty}\frac{ n+2 }{ (2n+1)*2n }=0 < 1\]
So by the Ratio Test the series converges.