Is the Basic Series Comparison Test Applicable for Sum(n^(1/2)/(n^2+1),n,1,inf)?

[nameVoid]

14. sum(n^(1/2)/(n^2+1),n,1,inf)

b=n^(1/2)/(n^2+2)<=n^(1/2)/(n^2+1)=a
b is conv since lim as n-> inf b = 0 since n^(1/2)<n^2+2 for x>=1 thus a is conv
i have a feeling this is a shakey way to do this if its even correct somone pleae clarify the solution to this problem

 

[Mark44]

Are you sure that n^1/2/(n² + 1) is convergent? This is not a standard series that someone would use for comparison.

Why not compare your series with one that is known to converge, namely n^1/2/n² = 1/n^3/2? This is a term from a convergent p-series.

 

[Elucidus]

I'm not completely positive that I understand your question.

Firstly I assume the series in question is \sum_{n=1}^\infty{\frac{\sqrt{n}}{n^2+1}}.

You seem to be making a couple of statements thereafter that I may have misinterpreted (please correct me if I am wrong). I believe your reasoning is incorrect.

A few comments. All of the following statements about series are FALSE!

\text{If }\lim_{n\rightarrow \infty}b_n=0\text{ then }\sum_{n=0}^\infty{b_n}\text{ converges}.

\text{If }0\leq b_n \leq a_n\text{ for all }n\text{ then }\sum_{n=0}^\infty{a_n}\text{ converges if }\sum_{n=0}^\infty{b_n}\text{ converges}.

\text{If }0\leq a_n&lt;b_n\text{ for all }n\text{ then }\sum_{n=0}^\infty{\frac{a_n}{b_n}}\text{ converges}.

This series can be shown to be convergent using the Comparison Test (or Limit Comparison Test) to an appropriate convergent p-series.

--Elucidus

 

[Mark44]

nameVoid is unclear in his question. The problem as stated lists
\sum_{n=1}^\infty{\frac{\sqrt{n}}{n^2+1}}.
but the following work also lists
\sum_{n=1}^\infty{\frac{\sqrt{n}}{n^2+2}}
in the comparison.

I think he wants to show that the first series above is convergent.

 

[Elucidus]

nameVoid is unclear in his question. The problem as stated lists
\sum_{n=1}^\infty{\frac{\sqrt{n}}{n^2+1}}.
but the following work also lists
\sum_{n=1}^\infty{\frac{\sqrt{n}}{n^2+2}}
in the comparison.

I think he wants to show that the first series above is convergent.

This is what I suspect. The reasoning that follows looks off and may have fallen to one of the non-facts in my previous post.

Your comment about comapring to \sum_{n=1}^\infty\frac{1}{n^{3/2}} is sound advice. The integral test would also work, but leads to a recalcitrant integral.

--Elucidus

 

[nameVoid]

the series converges because the second integral 1/n^(3/2) converges and is greater than first n^(1/2)/(n^2+1) correct me if I am wrong here but
if b converges and a<=b a must also converge
if b diverges and a>=b then a also must diverge

 

[Mark44]

the series converges because the second integral 1/n^(3/2) converges and is greater than first n^(1/2)/(n^2+1) correct me if I am wrong here but
if b converges and a<=b a must also converge
if b diverges and a>=b then a also must diverge

Yes, sort of, but you're leaving out some important details and mangling a few things. What you want to say is this:
a_n = n^1/2/(n² + 1) <= 1/n^3/2 = b_n for all n >= 1.
\sum b_n is a convergent p-series.[/itex]
Therefore \sum a_n converges as well.

For your last two statements, you are omitting some important information, such as that the series involved have to consist of nonnegative terms, and that you are talking about terms in infinite series.

 

[nameVoid]

thank you.
how about this next case
sum((2+cosn)/n^2,n,1,inf)
sum(2/n^2,n,1,inf)+sum(cosn/n^2,n,1,inf)>=1/n^2 which is a convergent p series
the book states here to use the basic comparison tests yet they tell me nothing here is there a way to chose a series such that it is greater than the first

 

[Mark44]

1 <= 2 + cos(n) <= 3 for all n

Can you find a convergent p-series whose terms are all larger than those of your series?

Hint: if \sum a_n is a convergent series, then so is \sum k a_n, where k is a constant. If the first series converges to a, the second converges to ka.

 

[nameVoid]

nice, (2+cosn)/n^2<3/n^2
3/n^2 is a convergent p series thus (2+cosn)/n^2 is also convergent

 

[Mark44]

Works for me!