Question

Can someone Please explain this problem to me?

Answer 1

\[\sum_{k=1}^{\infty}\frac{ k(k+2) }{ (k+6)^2 }\]

Answer 2

Apparently that Diverges.

But I dont see how. If you expand it all out, they exponents are the same. And the denominator is bigger.

Answer 3

@zepdrix @mathmate

Answer 4

@SolomonZelman @Directrix

Answer 5

Are you sure it diverges? It converges for sure.

Answer 6

Whoever or whatever said it diverges probably made a mistake.

Answer 7

The leading terms are k^2, right?
So as the terms get bigger and bigger, each term is approaching a value of 1 (I can explain this further if you're confused).

So you're adding an infinite number of 1's together: Divergent.

Answer 8

But isnt that just 1, so wouldn't that be convergent?
I thought any constant number would be convergent

Answer 9

Maybe you're confusing `sequence` with `series`.
Marceilli was making that mistake yesterday :D lol.

So if you look at the limit,\[\large\rm \lim_{n\to\infty}\frac{k(k+2)}{(k+6)^2}=\frac11\]This tells you that once we get to a really large nth term,
the term is 1... ish.

So then your big nth term is 1ish, \(\large\rm a_n=1\)
and all the terms after that are 1 also, right?

So you're adding a bunch of 1's together.

Answer 10

That adds up to an infinite amount.

The limit must give you zero for it to converge.

Answer 11

The `terms` are converging to 1, yes.
But the `sum` is diverging to infinity because you're adding up a bunch of 1's.

Answer 12

Oh, so it's basically just adding 1 infinite amount of times right?

Answer 13

Ya I think you silly billy's are thinking of the sequence D:
The sequence converges, yes.

Answer 14

But not the series.


Where you at Sid? :U confused?

Answer 15

No I think i get it now. But how do you distinguish a sequence problem from a series problem. I learned Sequences in section 1 and series in section 2. But the problems practically look the same to me D:

Answer 16

Oh is it that sequences dont have the summation symbol?

Answer 17

That's the thing, the wording will be almost identical.
You have to look for the keyword.

A sequence problem will look like this usually:
"Determine whether or not the sequence converges,
if it does, determine the limit value."

A series problem will look like this usually:
"Determine whether or not the series converges,
if it does, determine its sum"

Answer 18

But yes, summation is a series :)
Good call. Look for that.

Answer 19

Oh okay I see.

So for series, the summation has to = 0 for it to converge.
Any other number would mean that it diverges?

Answer 20

Usually for sequence they will put those weird squiggly brackets around it.\[\large\rm \left\{~\sin(x)e^{-x}~\right\}_{n=0}^{\infty}\]

Answer 21

It's interesting to see how they can take algebra 2 math and introduce it again in calc 2.

Answer 22

In order for the `series` to converge,
then the `sequence` must converge to zero, yes.

Think about it and it might seem logical.
When your n gets big enough, eventually the terms have to be 0's,
otherwise you're adding a bunch of non-zero stuff.

That doesn't guarantee convergence,
(Think about the harmonic series)
\(\large\rm \lim\limits_{n\to\infty}\frac1n=0\)
which we know is divergent.

But it at least gives you something to work with.

Answer 23

So I guess maybe think of it like this...

If the sequence converges to zero,
then the series has the potential to be convergent.

if the sequence converges to any other value,
then the series diverges.

Answer 24

Yes,

If your sequence does not converge to zero, series (definitely) diverges.

Answer 25

you can compare with the harmonic series,

\(\displaystyle \sum_{k=j}^{\infty}\frac{ k(k+2) }{ (k+6)^2 }\ge\sum_{k=j}^{\infty}\frac{ (k+6)}{ (k+6)^2}=\sum_{k=j+6}^{\infty}\frac{ 1}{ k}\)

if you really like ((it's a joke))

Answer 26

Sorry. I am getting confused again.

I thought sequences and series where different problems. How are you looking at the sequence of a series problem?

Answer 27

Suppose you have any infinite series (will call it S).
Then, suppose that the sequence converses to x.

So, at some point (after some Nth term), you are (pretty much) adding x infinitely many times. So, if this x is NOT ZERO, than the series diverges.

Answer 28

agree with this logic?

Answer 29

Oh yes. That makes sense

Answer 30

Sequence is the list of terms.
\(\large\rm a_1, a_2, ...,a_n\)

Series is the sum of those terms.
\(\large\rm a_1+a_2+...+a_n\)

Answer 31

Do you see how they relate?

Answer 32

So a sequence basically looks at it by 1 term
While a series looks at multiple together

Answer 33

ya sure, that's a way to think of it :)

Answer 34

\(\dfrac{k^2+2k+1-1}{(k+6)^2}=\dfrac{(k+1)^2}{(k+6)^2}-\dfrac{1}{(k+6)^2}\)

As \(k\rightarrow \infty\), the first term \(\rightarrow 1\), the second term \(\rightarrow 0\)

Hence, it is converge. To me : )

Answer 35

I thought it was Divergent no?

Answer 36

Wait so if the exponents of a given summation of a series is the same, it will go to 1 and Diverge. True?

Answer 37

as n --> infinity,
A_n = 1, right?

so you are adding 1's infinitely many times. Yes?

Answer 38

Yes, polynomial of nth degree / polynomial of nth degree, then sequence converges to 1, and series diverges.

Answer 39

Oh okay i see it now. Thanks guys! I really appreciate the help! :3

Answer 40

Yayyy

Answer 41

Wait another question.

Answer 42

So it would work the same way if the exponent was "n"

\[\sum_{n=1}^{\infty} \frac{1+8^n }{ 5^n }\]

Answer 43

Since the exponents are the same. The sequence converges to 1. Thus the summation diverges. Right?

Answer 44

Exponentials? Naw this is a bit different.
That rule applies to polynomials.\[\large\rm \lim_{n\to\infty}\frac{1+8^n}{5^n}\quad=\quad \lim_{n\to\infty}\left(\frac15\right)^n+\left(\frac85\right)^n\]

8/5 is larger than 1,
so when you multiply it by itself,
it's going to keep getting bigger, right?
(8/5)^(7000) = 3 bajillion.

Answer 45

It diverges, yes,
but not for the reason you thought.

Answer 46

Oh I see.

Answer 47

Thanks. I understand it now!