convergance of series

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May 18th 2006, 12:05 PM
Aradesh

convergance of series

Show that if $\sum_{r=1}^{\infty} u_r$ converges, then:

$\sum_{r=1}^{\infty} \frac{r+10}{r} u_r$ also converges.
May 18th 2006, 12:57 PM
topsquark

Quote:

Originally Posted by Aradesh

Show that if $\sum_{r=1}^{\infty} u_r$ converges, then:

$\sum_{r=1}^{\infty} \frac{r+10}{r} u_r$ also converges.

For large r, the second series approaches the first (ie. $\lim_{r \to \infty} \frac{r+10}{r}u_r \to \lim_{r \to \infty}u_r$ . For large r, the coefficient approaches 1.) Thus the second series will converge since the first does.

If you need specifics you can get a bit fancier and do a ratio test:
$\lim_{r \to \infty} \frac{ \frac{(r+1)+10}{r+1}u_{r+1} } { \frac{r+10}{r}u_r }$ = $\lim_{r \to \infty} \frac{r(r+11)}{(r+1)(r+10)} \frac{u_{r+1}}{u_r}$ = $\lim_{r \to \infty} \frac{u_{r+1}}{u_r}$

Since the series $\sum_{r=1}^{\infty}u_r$ converges we know that $\lim_{r \to \infty} \frac{u_{r+1}}{u_r} < 1$ . Thus the series $\sum_{r=1}^{\infty}\frac{r+10}{r}u_r$ must also converge.

-Dan
May 18th 2006, 05:23 PM
Aradesh

"For large r, the second series approaches the first" ... "Thus the second series will converge since the first does."

can you prove this? ie that.. if the terms of a second series approach the terms of a first converging series, then the second series converges also?

and as to the ratio test, i never said anything about $\frac{u_{n+1}}{u_n}<k<1$ surely there could be convergent series that do not satisfy this?
May 18th 2006, 05:55 PM
ThePerfectHacker

Quote:

Originally Posted by Aradesh

Show that if $\sum_{r=1}^{\infty} u_r$ converges, then:

$\sum_{r=1}^{\infty} \frac{r+10}{r} u_r$ also converges.

Assuming, $u_r>0$

You have more simply,
$\sum_{r=1}^{\infty} (1+10/r) u_r=\sum_{r=1}^{\infty} u_r+\frac{10 u_r}{r}$
Note that $\sum_{r=1}^{\infty} u_r$ congerves.
Then,
$\sum_{r=1}^{\infty}\frac{u_r}{r}$ must converge because,
$0\leq \frac{u_r}{r} \leq u_r$ -this is direct comparassion test.
Then so too,
$\sum_{r=1}^{\infty}\frac{10u_r}{r}$ must converge because all you are doing is multipling and infinite series by a constant does not change its convergence.
Since the two summands of,
$\sum_{r=1}^{\infty} u_r+\frac{10 u_r}{r}$
converges, implies that this sum must converge.

Q.E.D.
May 18th 2006, 06:01 PM
ThePerfectHacker

Quote:

Originally Posted by topsquark

Since the series $\sum_{r=1}^{\infty}u_r$ converges we know that $\lim_{r \to \infty} \frac{u_{r+1}}{u_r} < 1$ .

I believe that the ratio test states that is the ratio converges to less than one then the series must converge.

If the series converges does not mean that the ratio test implies it being less than one. It cannot imply more than one for that would imply that is diverges. However it can still imply that is precisely equal to 1, (remember ratio test inconclusive at L=1).
May 19th 2006, 03:27 AM
topsquark

Quote:

Originally Posted by ThePerfectHacker

I believe that the ratio test states that is the ratio converges to less than one then the series must converge.

If the series converges does not mean that the ratio test implies it being less than one. It cannot imply more than one for that would imply that is diverges. However it can still imply that is precisely equal to 1, (remember ratio test inconclusive at L=1).

:o (Ahem!) Forgot about that little point. Sorry!

-Dan
May 19th 2006, 08:47 AM
Aradesh

I believe it, but i'm not convinced by one of your steps, tph.

ie that:
$\sum_{r=1}^{\infty} \frac{u_r}{r}$ must converge since

$0 \le \frac{u_r}{r} \le u_r$

Aren't you are using this to say that since the terms in the series $\frac{u_1}{1} + \frac{u_2}{2} + \frac{u_3}{3} + \cdots$ are less than or equal to each corresponding term in the series $u_1 + u_2 + u_3 + \cdots$ , therefore the first series must converge since the second one does.
But as far as I know, you can only say that if you know all the terms in the series are either all positive, or all negative (for sufficiently large r). since you'd know, for example, that the increasing smaller series would be bound above by the limit of the series you are using to compare with. but if we don't know that the series might be alternating, or changing signs erratically, I don't think you can reason it like this.

so could you elaborate on the fact that since
$0< |\frac{u_r}{r}| < |u_r|$
then
$\sum_{r=1}^{\infty}u_r \mbox{ converges } \Longrightarrow \sum_{r=1}^{\infty}\frac{u_r}{r} \mbox{ converges }$
?

thanks.
May 19th 2006, 09:56 AM
ThePerfectHacker

Quote:

Originally Posted by Aradesh

so could you elaborate on the fact that since
$0< |\frac{u_r}{r}| < |u_r|$
then
$\sum_{r=1}^{\infty}u_r \mbox{ converges } \Longrightarrow \sum_{r=1}^{\infty}\frac{u_r}{r} \mbox{ converges }$
?

thanks.

If the series is absolutely convergent, how do you know? If the problem says it is then you can state that.
May 23rd 2006, 05:34 PM
Aradesh

Let $k$ be a constant such that:
$k>|U_n|$ for all positive integer n.
I believe this is another problem.
How do you know that the inequality that follows this proposition is still valid.

Yes, $k>|U_n|$ , but how do you know that,
$k>|U_{n+1}|$ therefore the inequality (on which you base your proof) is imprecise.
May 23rd 2006, 06:51 PM
ThePerfectHacker

Quote:

Originally Posted by Aradesh

$0 < |\sum_{r=1}^\infty \frac{u_r}{r}| < k$
since k is finite, the series converges. $\blacksquare$

Maybe I am missing something but, just because,
$0 < \left| \sum_{r=1}^n \frac{u_r}{r} \right| < k$
Does not mean that is converges,

Indeed,
$0<\sum_{k=1}^{n}(-1)^k<2$
But it does not converge :eek:
May 24th 2006, 02:16 PM
Aradesh

you're right. so i've changed it a bit. what do you think now? is it okay?
May 24th 2006, 02:40 PM
ThePerfectHacker

I am so sorry!
I wanted to QUOTE repley but I accidently pressed EDIT and my moderator priveleges deleted your entire post!!

Your proof is gone :(,

Anyway I believe it was wrong because,

Quote:

Originally Posted by Aradesh

Let $k$ be a constant such that:
$k>|U_n|$ for all positive integer n.

I believe this is another problem.
How do you know that the inequality that follows this proposition is still valid.

Yes, $k>|U_n|$ , but how do you know that,
$k>|U_{n+1}|$ therefore the inequality (on which you base your proof) is imprecise.
May 24th 2006, 03:19 PM
Aradesh

Well i meant that $U_n$ is bounded so there exists a $k$ , independent of n, such that for all $U_n$ the inequality $k > |U_n|$ is valid.
May 24th 2006, 06:15 PM
ThePerfectHacker

Quote:

Originally Posted by Aradesh

Well i meant that $U_n$ is bounded so there exists a $k$ , independent of n, such that for all $U_n$ the inequality $k > |U_n|$ is valid.

You should have show that,
$\forall n\in\mathbb{Z}^+,k>|U_n|$