Does This Infinite Series Converge to 1/(ln2√2)?

[NewScientist]

Interesting question - Not homework!

Show that the sum of the infinite series:
$$log_2 e - log_4 e + log_{16} e + {(-1^n)} log_{2^{2n}} e ...$$
equals $$\frac{1}{ln2\sqrt2}$$
Any ideas?!

 

[matt grime]

what is the relationship between log base A and log base B? converting all those logs into base e would be the place to start

 

[roger]

are you sure $${(-1^n)} log_{2^{2n}} e ...$$ is correct ?

 

[NewScientist]

Well perhaps there should be ... before it but stilll...any ideas?! on how to proove?

 

[matt grime]

Yes, I told you, change base in logs so that they are all base e.

 

[mezarashi]

matt has been referring to the identity: $$log_2 e = \frac{ln e}{ln 2}$$


$$log_2 e - log_4 e + log_{16} e + ... {(-1)^n} log_{2^{2n}} e$$

$$\frac{lne}{ln2} - \frac{lne}{ln4} + ... {(-1)^n} \frac{lne}{ln2^n}$$


$$\frac{1}{ln2} ( 1 - \frac{1}{2} + ... {(-1)^n} \frac{1}{2^n})$$


Geometric series formula still at hand?

 

[shmoe]

Your nth term can't be correct. Do you know the term right after $$\log_{16} e$$?

 

[roger]

mezarashis nth term isn't correct either

 

[dx]

The nth term is $$(-1)^{n+1}log_{2^n}{e}$$

 

[shmoe]

The nth term is $$(-1)^{n+1}log_{2^n}{e}$$

That won't sum to $$\frac{1}{\log 2^{3/2}}$$ either, and it doesn't match the 3rd term of the given series.

For that matter the nth term NewScientist supplied doesn't match the first term. No need to look to the fourth to see it's not correct, though the 4th should give more insight on what the nth term is.

 

[NewScientist]

No dx, the nth term is
$$log_{2^{2n}} e$$
The step : $$\frac{lne}{ln2} - \frac{lne}{ln4} + ... {(-1)^n} \frac{lne}{ln2^n}$$
is correct
Then one must obsewrve this is a GP with r = -1/2 and a 1/In 2^2n
Sum to infinity using a / (1-r)
Which gives
$$\frac{\frac{1}{ln2}}{1+\frac{1}{2}}$$
This simplifies to
$$\frac{1}{1.5 ln2}$$
$${1.5 ln2} = ln 2^\frac{3}{2}$$
which gives
$$ln 2\sqrt{2}$$

 

[shmoe]

No dx, the nth term is
$$log_{2^{2n}} e$$

If this is your nth term then:

The step : $$\frac{lne}{ln2} - \frac{lne}{ln4} + ... {(-1)^n} \frac{lne}{ln2^n}$$
is correct

is wrong. $$\log_{2^{2n}} e=\frac{\log e}{\log 2^{2n}}$$

In any case, the sum that you wrote that I just quoted is Not a geometric series. It's a constant times $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$

 

[roger]

If this is your nth term then:



is wrong. $$\log_{2^{2n}} e=\frac{\log e}{\log 2^{2n}}$$

In any case, the sum that you wrote that I just quoted is Not a geometric series. It's a constant times $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$

but the first term, n=1 is positive though

 

[shmoe]

but the first term, n=1 is positive though

"a constant times"

 

[HallsofIvy]

No dx, the nth term is
$$log_{2^{2n}} e$$
The step : $$\frac{lne}{ln2} - \frac{lne}{ln4} + ... {(-1)^n} \frac{lne}{ln2^{2n}}$$
is correct
Then one must obsewrve this is a GP with r = -1/2 and a 1/In 2^2n

No, it isn't. For one thing, if the general term is $log_{2^{2n}} e$ as you keep insisting, then the first term, with n= 0 would be $log_1 e$ which does not exist. If we take n= 1 as the first term then it would be $log_{4} e$, not $log_{2} e$. However, let's assume that the general form is $log_{2^{2n}} e$, starting with n= 1 and add on $log_2 e$ at the end.
Since ln e= 1 (I don't know why everyone kept writing ln e.) This series is
[tex]\frac{1}{2 ln 2}\Sigma_{n=1}^\infnty \frac{(-1)^n}{n}[/itex]<br /> except for the factor $\frac{1}{2 ln 2}$ this is the "alternating harmonic series" which is well known to converge to ln 2. In other words that series converges to 1/2! Adding on that first term, $log_2 2= \frac{1}{ln 2}$[/tex][itex]which did not conform to the given general term, we have <br /> $$\frac{1}{2}+ \frac{1}{ln 2}= \frac{1+ ln 2}{2 ln 2}$$[/itex]

 

[NewScientist]

Okay, I have been jumping backwards and forwards to this question all day so here is my full solution - one i have thought about properly! I hope I can make some of you eat your words!
Lets go
$$log_2 e - log_4 e + log_{16} e + ... + {(-1^n)} log_{2^{2n}} e ...$$
This is given in the question and yes the 'fourth' term (the general term IS correct, let n= 0,1,2,3...
Now, the 'end result' is in ln so let's convert to base e and see where thatgets us
$$log_a x = \frac{log_b x}{log_b a}$$
Let
$$x = e = b$$
$$a = 2^{2n}$$

This gives

$$log_2 e - log_4 e + log_{16} e + ... + {(-1^n)} log_{2^{2n}} e ... = \frac{ln e}{ln 2} - \frac{ln e}{ln 4} + ... + {(-1^n)} \frac{ln e}{ln 2^{2n}}$$

$$log_x x = 1 \rightarrow lne = 1$$

Which gives

$$log_2 e - log_4 e + log_{16} e + ... + {(-1^n)} log_{2^{2n}} e ... = \frac{1}{ln 2} - \frac{1}{ln 4} + ... + \frac{(-1^n)}{ln 2^{2n}}$$

It is clear that the common ratio of the terms is -1/2. The first term is 1/ln2. The series is infinite so n = infinity! Let us sum this GP.

$$Sn = \frac{a}{1-r}$$

$$\frac{\frac{1}{ln2}}{1+\frac{1}{2}}$$

$$\frac{1}{1.5 ln2}$$

$${1.5 ln2} = ln 2^\frac{3}{2}$$

$$ln 2\sqrt{2}$$

QED

 

[shmoe]

Lets go
$$log_2 e - log_4 e + log_{16} e + ... + {(-1^n)} log_{2^{2n}} e ...$$
This is given in the question and yes the 'fourth' term (the general term IS correct, let n= 0,1,2,3...

As Halls pointed out, n=0 in your general term does not exist and hence is not $$\log_2 e$$.

Try $$(-1)^n \log_{2^{(2^n)}} e$$ as your general term, starting at n=0.

edit- again, even using your general term you DO NOT get a geometric series. With your general term your n=2 and n=3 terms are $$\frac{1}{\log{16}}-\frac{1}{\log{64}}=\frac{1}{4\log{2}}-\frac{1}{6\log{2}}$$... the n=3 term is not -1/2 times the n=2 term.

 

[NewScientist]

n can = 0

edit : $$\frac{1}{ln2} = 2\frac{1}{ln4}$$

thus we have a geometric series - don't we?!

-NS

 

[shmoe]

n can = 0

-NS

$$\log_{2^{2(0)}}e=\log_{2^0}e=\log_{1}e$$ which is undefined, unless you can tell me what power to raise 1 to to get e...

 

[NewScientist]

How do i correct proof then?! As my result is correct?! And indeed i never claimed log of base 2^2n worked for the first term :P

 

[shmoe]

edit : $$\frac{1}{ln2} = 2\frac{1}{ln4}$$

thus we have a geometric series - don't we?!

-NS

Look at more terms that your proposed general term will give, it will not be a geometric series and it will not sum to what you claimed the answer was (Halls worked it out for you, also taking into account the mismatched first term).


Try the general term I proposed above.

 

[NewScientist]

Thankyou - and sorry! The STEP paper had that general formula and that flumoxed me!

However, Halls answer does not agree with the paper - who is correct? I'd back the Cambridge guys that set the paper!

 

[shmoe]

I would say there's a typo in the paper or how you are reading it. compare:

$$(-1)^n \log_{2^{2^n}} e$$

and

$$(-1)^n \log_{2^{2n}} e$$

As the size of the type used for the base gets smaller the probability of misreading it goes up. With $$2^{2^n}$$ in the base, you get the claimed sum and the terms all match.

 

[shmoe]

So is my result correct?!

You summed the wrong series using a method that didn't apply , so I'm going to have to say no.

Try again with the new version. You will get a geometric series in this case, and everything will work out

 

[NewScientist]

A sequence with r = -1/2 ?!

 

[matt grime]

what's wrong with that?

 

[NewScientist]

Nothing! It has just dawned on me that because I copy and paste TEX through my working that the 2^(2n) was more than likely a mistype otherwise I would not have arrived at my solution!

 

[HallsofIvy]

A couple of hours after my post (when I was not near a computer!) it suddenly dawned on me that NewScientist must have meant
$$\Sigma log_{2^{2^n}} e$$

That reduces to
$$\Sigma \frac{(-1)^n}{ln(2^{2^n}}= \Sigma \frac{(-1)^n}{2^n ln 2}= \frac{1}{ln 2}\Sigma \left(\frac{-1}{2}\right)^n$$
which is, in fact, a simple geometric series.

 

[benorin]

That's Beautiful.

A couple of hours after my post (when I was not near a computer!) it suddenly dawned on me that NewScientist must have meant

$$\Sigma log_{2^{2^n}} e$$

That reduces to

$$\Sigma \frac{(-1)^n}{ln 2^{2^n} }= \Sigma \frac{(-1)^n}{2^n ln 2}= \frac{1}{ln 2}\Sigma \left(-\frac{1}{2}\right) ^n$$
which is, in fact, a simple geometric series.