Infinity minus Infinity Question

[RJLiberator]

For each series, decide whether they converge or not:
k≥1∑2/(k(k+2))

So, I am trying to use the Integral test on this problem. After applying partial fractions I receive the answer of ln(x)-ln(x+2) from 1 to infinity.

Evaluating 1 is pretty straight forward.. ln(3). Which seems to be the correct answer.

Infinity, however, leaves me with ln(infinity) - ln(infinity). My initial thinking is that this is equal to 0 and all is well.

However, is it OK to evaluate this in such a manner? Infinity-infinity=0 ??
Or is that mechanically wrong?

 

[haruspex]

For each series, decide whether they converge or not:
k≥1∑2/(k(k+2))

So, I am trying to use the Integral test on this problem. After applying partial fractions I receive the answer of ln(x)-ln(x+2) from 1 to infinity.

Evaluating 1 is pretty straight forward.. ln(3). Which seems to be the correct answer.

Infinity, however, leaves me with ln(infinity) - ln(infinity). My initial thinking is that this is equal to 0 and all is well.

However, is it OK to evaluate this in such a manner? Infinity-infinity=0 ??
Or is that mechanically wrong?

You certainly cannot subtract infinity from infinity (or divide infinity by infinity) and get a meaningful answer.
In the way you have solved the problem you have effectively rearranged a convergent series into two divergent ones. Consider this sequence:
1 + 1/2 + 1/4 + ...
Converges no problem.
Now rewrite it as (1-0) + (1-1/2) + (1-3/4) ... and rearrange as (1+1+1...) - (0+1/2+3/4+...). You see what has happened?
On the other hand, you are allowed to sum groups of adjacent terms of a convergent series without changing the order. 1+1/2+1/4+... can be processed (if it helps) as (1+1/2)+(1/4+1/8)+... After splitting up your original terms using partial fractions, can you see how to apply that?

 

[gopher_p]

Use log laws to rewrite the expression $$\ln x-\ln(x+2)=\ln\frac{x}{x+2}$$ and then take the limit.

 

[RJLiberator]

Beautiful. Well understood and makes much more sense. So the limit turns out to be 1.

Does this mean my final answer is 1+ln(3) ?
This seems incorrect as when I plug in the initial integral into calculators the answer is just ln(3)

:/ Or is this just off by a constant?

 

[gopher_p]

Beautiful. Well understood and makes much more sense. So the limit turns out to be 1.

Does this mean my final answer is 1+ln(3) ?
This seems incorrect as when I plug in the initial integral into calculators the answer is just ln(3)

:/ Or is this just off by a constant?

The limit should be $\lim\limits_{x\rightarrow\infty}\ln\frac{x}{x+1}=\ln 1=0$.

 

[RJLiberator]

Ah, I see. Thank you kindly.

 

[RJLiberator]

You certainly cannot subtract infinity from infinity (or divide infinity by infinity) and get a meaningful answer.
In the way you have solved the problem you have effectively rearranged a convergent series into two divergent ones. Consider this sequence:
1 + 1/2 + 1/4 + ...
Converges no problem.
Now rewrite it as (1-0) + (1-1/2) + (1-3/4) ... and rearrange as (1+1+1...) - (0+1/2+3/4+...). You see what has happened?
On the other hand, you are allowed to sum groups of adjacent terms of a convergent series without changing the order. 1+1/2+1/4+... can be processed (if it helps) as (1+1/2)+(1/4+1/8)+... After splitting up your original terms using partial fractions, can you see how to apply that?

Haruspex, I am trying to decipher your quote. Are you saying that I made a mistake by starting off with partial fractions and thus corrupting the problem?

 

[LCKurtz]

Have you thought about a simple comparison test with $\sum\frac 2 {k^2}$?

 

[RJLiberator]

Have you thought about a simple comparison test with $\sum\frac 2 {k^2}$?

Ha..Ha... I... see what you did there.

Brilliant. Ah well, I am glad I am working out more then one way, but THIS way is by far and away the most simple and what the assignment was going for. Thank you for bringing it to my attention.

 

[haruspex]

Haruspex, I am trying to decipher your quote. Are you saying that I made a mistake by starting off with partial fractions and thus corrupting the problem?

No, partial fractions was an excellent start. But instead of the extracting each fraction form into a separate series, you do better merely to regroup the terms at a more local level. Write out the first, say, 4 terms of your pair of partial fractions. I.e. 1st term first fraction, 1st term second fraction, etc. Notice anything?

 

[RJLiberator]

After writing them out, there is a very clear decreasing pattern.
It seems to be:
(1-1/3)(1/2-1/4)(1/3-1/5)...
(1/n-1/(n+2))

hm, It appears there is some common pattern here that I am unfamiliar with and thus, it is not popping out to me.

 

[haruspex]

(1-1/3)(1/2-1/4)(1/3-1/5)...

Why are you multiplying the terms together. It's a sum, no?

 

[RJLiberator]

Oh, I mean addition, not multiplication.
This appears to be some form of a telescoping series.

It is not clearly visible to me because I am used to the very basic telescoping series.

This one seems to have 1+1/2 left over in the beginning and it must have some form of an n value left at the end.
Hm.

 

[haruspex]

This one seems to have 1+1/2 left over in the beginning

Right.

some form of an n value left at the end.

.. which in the limit tend to what?

 

[RJLiberator]

Hm. The limit would have them going to 0.

So, 1+1/2 seems to be the sum. Is that correct?

 

[haruspex]

Hm. The limit would have them going to 0.

So, 1+1/2 seems to be the sum. Is that correct?

Yes!

 

[RJLiberator]

Ah, absolutely beautiful. The telescoping series is quickly becoming one of my favorite series. It seems to appear due to partial fractions.

If anyone is checking this thread:

Is the sum from 0 to infinity of (k^(1/2)+2)/(k+5) in perfect form for the p-series test? p = 1/2 or is this NOT true due to the addition of 2 and 5.

 

[haruspex]

Is the sum from 0 to infinity of (k^(1/2)+2)/(k+5) in perfect form for the p-series test? p = 1/2 or is this NOT true due to the addition of 2 and 5.

It is a forum standard that a new question requires a new thread ;)

 

[RJLiberator]

Np, thanks haruspex for all of your insight. I appreciate it, greatly.