Series Converge Or Diverge?

[dagr8est]

Does the series given by:

first term = 1
(n+1)th term = n(n+1)(nth term)/[(n+2)(n+3)]

converge or diverge?

I tried to find a way to write this using sigma notation without the recursion but no luck. The question says to generalize the pattern first so I think there is a way to write it in sigma notation and then apply one of the various converge/diverge tests but I can't figure it out. Any help would be appreciated.

 

[soroban]

Hello, dagr8est!

I tried to write it out, but it got very unwieldy . . .

. . \(\displaystyle 1 \,+\,(1)\left(\frac{2\cdot3}{4\cdot5}\right) \,+\,(1)\left(\frac{2\cdot3}{4\cdot5}\right)\left(\frac{3\cdot4}{5\cdot6}\right)\,+\,(1)\left(\frac{2\cdot3}{4\cdot5}\right)\left(\frac{3\cdot4}{5\cdot6}\right)\left(\frac{4\cdot5}{6\cdot6}\right) \,+\,\cdots\)

Some cancelling was possible, but I saw no discernible pattern.

Converge or diverge?

\(\displaystyle a_1\,=\,1,\;\;a_{n+1}\:=\:\frac{n(n+1)}{(n+2)(n+3)}\,a_n\)

I would use the Ratio Test . . .

\(\displaystyle \L R \:=\:\frac{a_{n+1}}{a_n} \;= \;\frac{\frac{n(n+1)}{(n+2_n+3)}\,a_n}{a_n} \;=\;\frac{n(n+1)}{(n+2)(n+3)} \;=\;\frac{n^2\,+\,n}{n^2\,+\,5n\,+\,6}\)


Divide top and bottom by \(\displaystyle n^2:\L\;\frac{\frac{n^2}{n^2}\,+\,\frac{n}{n^2}}{\frac{n^2}{n^2}\,+\,\frac{5n}{n^2}\,+\,\frac{6}{n^2}} \;= \;\frac{1\,+\,\frac{1}{n}}{1\,+\,\frac{5}{n}\,+\,\frac{6}{n^2}}\)

Then: \(\displaystyle \L\:\lim_{n\to\infty}R\;=\;\lim_{n\to\infty}\left[\frac{1\,+\,\frac{1}{n}}{1\,+\,\frac{5}{n}\,+\,\frac{6}{n^2}}\right] \;= \;\frac{1\,+\,0}{1\,+\,0\,+\,0}\;=\;1\)

. . Therefore, the series diverges.

 

[dagr8est]

I also thought about using the ratio test but it is inconclusive if the result is 1. If <1, then the series converges or if >1, the series diverges.

 

[pka]

You are of course correct! The limit 1 is inclusive: a common mistake.
However, has Soroban read your problem correctly?
Frankly, I am not sure of the definition of the terms in the series.

Do you have any advanced tests? Do you know Kummer’s test?
From which the Gauss test can be derived.

 

[dagr8est]

Soroban, I think the second term should be 1(2)/[3(4)] because n=1 for term a(2).

Now that I look at it more, I think the sigma notation form is sigma(n=1->infinity) 12n!(n-1)!/[(n+2)!(n+1)!] = sigma(n=1->infinity) 12/[n(n+1)^2(n+2)]. So if that is compared to 12/(n^4) which is always greater, the series converges.

 

[pka]

Will you post the question as nearly as it is written as possible?
Because \(\displaystyle \frac{{12n!(n - 1)!}}{{\left( {n + 2} \right)!\left( {n + 1} \right)!}} = \frac{{12}}{{\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n + 1} \right)!}}.\)

That sum converges!

 

[dagr8est]

Will you post the question as nearly as it is written as possible?
Because \(\displaystyle \frac{{12n!(n - 1)!}}{{\left( {n + 2} \right)!\left( {n + 1} \right)!}} = \frac{{12}}{{\left( {n + 2} \right)\left( {n + 1} \right)\left( n \right)\left( {n + 1} \right)!}}.\)

That sum converges!

The original question is:

Does this series converge or diverge?

first term = 1
(n+1)th term = n(n+1)(nth term)/[(n+2)(n+3)]

The other stuff I wrote in the previous post was just me trying to write the problem using sigma notation without recursion.

 

[pka]

If this the way it is written:
\(\displaystyle \L a_1 = 1\quad \& \quad a_{n + 1} = \frac{{n\left( {n + 1} \right)}}{{\left( {n + 2} \right)\left( {n + 3} \right)}}a_n ?\)

 

[dagr8est]

If this the way it is written:
\(\displaystyle \L a_1 = 1\quad \& \quad a_{n + 1} = \frac{{n\left( {n + 1} \right)}}{{\left( {n + 2} \right)\left( {n + 3} \right)}}a_n ?\)

yes

 

[pka]

It would really help if you would learn some basic TeX.

It does converge and the sum is approx 1.261.

But its proof depends on what tests you know.