# Interesting question - Not homework

1. Nov 24, 2005

### 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?!

Last edited: Nov 24, 2005
2. Nov 24, 2005

### 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

3. Nov 24, 2005

### roger

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

4. Nov 24, 2005

### NewScientist

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

5. Nov 24, 2005

### matt grime

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

6. Nov 24, 2005

### 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?

Last edited: Nov 24, 2005
7. Nov 24, 2005

### shmoe

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

8. Nov 24, 2005

### roger

mezarashis nth term isnt correct either

9. Nov 24, 2005

### dx

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

10. Nov 24, 2005

### shmoe

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.

Last edited: Nov 24, 2005
11. Nov 24, 2005

### 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
$$ln 2\sqrt{2}$$

12. Nov 24, 2005

### shmoe

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 $$\sum_{n=1}^{\infty}\frac{(-1)^n}{n}$$

Last edited: Nov 24, 2005
13. Nov 24, 2005

### roger

but the first term, n=1 is positive though

14. Nov 24, 2005

### shmoe

"a constant times"

15. Nov 24, 2005

### HallsofIvy

Staff Emeritus
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
$$\frac{1}{2 ln 2}\Sigma_{n=1}^\infnty \frac{(-1)^n}{n}[/itex] except for the factor $\frac{1}{2 ln 2}$ this is the "alternating harmonic series" which is well known to converge to ln 2. In other words that series converges to 1/2! Adding on that first term, [itex]log_2 2= \frac{1}{ln 2}$$ which did not conform to the given general term, we have
$$\frac{1}{2}+ \frac{1}{ln 2}= \frac{1+ ln 2}{2 ln 2}$$

Last edited: Nov 24, 2005
16. Nov 24, 2005

### 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 lets 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

Last edited: Nov 24, 2005
17. Nov 24, 2005

### shmoe

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.

Last edited: Nov 24, 2005
18. Nov 24, 2005

### NewScientist

n can = 0

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

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

-NS

Last edited: Nov 24, 2005
19. Nov 24, 2005

### shmoe

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

20. Nov 24, 2005

### 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