Question

If we know....

Answer 1

\[\sum_{n=0}^{\infty}a _{n}\] Converges

Answer 2

What, (if anything) can be known about:
\[\sum_{n=0}^{\infty}(1/a _{n})\]

Answer 3

Hi!

Answer 4

So converge means it sums up to a number

Answer 5

right.

Answer 6

But that number could also be zero so the statement that the 2nd summation also adds up to a real number is not true

Answer 7

hmm...this is a little confusing to me lol. a_n could converge to 0, but the series could converge to something else. For example, lets say:
\[a_n = (\frac{1}{2})^n\]
The sequence converges to 0, but the series converges to:
\[\frac{1}{1-\frac{1}{2}} = 2\]

Answer 8

You confusing what sequence converging vs what series converges to

Answer 9

so in that example:
\[\frac{1}{a_n} = \frac{1}{(\frac{1}{2})^n} = 2^n\]
and we know this sequence and series doesnt converge.

Answer 10

From my understanding the only way the series will converge is if a_n is just a constant or if it is/simplifies to a form similar to 1/n (convergence to 0)

Answer 11

i guess if \[\sum_{n=0}^{\infty}a_n\]
converges, we dont know anything about the other summation.

Answer 12

if a_n is constant, the series wont converge:
\[\sum_{n=0}^{\infty}1 = 1+1+1+1+1+1+\cdots\]

Answer 13

The sequence converges, but not the series.

Answer 14

ok

Answer 15

You're definitely confusing convergence of sequences and convergence of series. A sequence converges when \[\lim_{n \rightarrow \infty}a_{n}=L\] For some L.

A series converges when the series equals a number. So \[\sum_{n=0}^{\infty}a_{n}=L\]
For some L, when that happens the sequence converges to 0. However there are also sequences that converge to 0 but whose series diverge.

Answer 16

So back to the question. \[\sum_{n=0}^{\infty}a _{n}\] converges so \[\lim_{n \rightarrow \infty}a_{n}=0\]
and thus \[\lim_{n \rightarrow \infty}\frac{1}{a_{n}}=\infty\]
which gives us the result:
\[\sum_{n=0}^{\infty}\frac{1}{a _{n}}\]
diverges.

Answer 17

hmmm

Answer 18

Ok good to note the specification between the two

Answer 19

But I think what you're saying is sub consciously what I was thinking

Answer 20

lim a_n=0 means any number put in for n will yield 0, so because of that we know the lim 1/(a_n) will always yield 1/0

Answer 21

Which also means it is divergent

Answer 22

lim a_n=0 means any number put in for n will yield 0

that's not true. for example \[\lim_{n \rightarrow \infty}\frac{1}{n}=0\]
but \[\frac{1}{n}\neq 0\]
for all n.

Answer 23

hmm

Answer 24

So how are you concluding it is divergent?

Answer 25

Do you agree with this:
\[\lim_{n \rightarrow \infty}\frac{1}{a_{n}}=\infty\]

Answer 26

\[\lim_{n \rightarrow \infty}\frac{1}{a_{n}}=0\]
if the series where to converge. But apparently that's not true.

Answer 27

It would depend on what a_n is

Answer 28

if a_n is n then as n -> infin. then yes L = 0

Answer 29

@ Thomas9 no I don't agree with that statement.

that \[\lim_{n \rightarrow \infty}\frac{1}{a_{n}}=\infty\]

Answer 30

consider the sequence \[a_n=\frac{(-1)^n}{n^2}\]

clearly \[\lim_{n\to\infty}a_n=0\]

but \[\lim_{n\to \infty}\frac{1}{a_n}\] does not exist. It is not infinity.

the important thing here is that the limit is not zero...that's all we need for the sum to diverge.

Answer 31

I follow there

Answer 32

So could this statement be a suitable answer in your opinion: "Becuase series 1 is convergent, lim of a_n as n->inf. = 0"

Answer 33

"Since lim of a_n as n->inf. = 0, then we also know lim of 1/a_n as n->inf. = 0"

Answer 34

* = DNE (not 0 for the second.)

Answer 35

"Which allows us to conclude that the 2nd series is divergent"

Answer 36

Ill write it out with equations..

Answer 37

Zarkon is right, I went a little too fast there. Also your argument is fine like this mwmnj.

Answer 38

Because: \[\sum_{n=0}^{\infty}a _{n}\] Converges, we know:
\[\lim_{n \rightarrow \infty}a _{n}=0\]
Which also make this the following true:
\[\lim_{n \rightarrow \infty}(1/a _{n})=DNE\]
Which means:
\[\sum_{n=0}^{\infty}(1/a _{n})\]
is divergent

Answer 39

no...

\[\lim_{n \rightarrow \infty}(1/a _{n})\]

might be infinity...or -infinity...or maybe it just doesn't exist at all

all we care about is that the limit is not zero..thus this sum cannot converge