6

I'm a bit puzzled with the following:

  • $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=1$

  • $\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}=1$

Which essentially yields the identity:

$$\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}$$

Now obviously, $\displaystyle\forall{n\in\mathbb{N}}:\frac{1}{2^n}\neq\frac{1}{2^n\ln(2^n)}$

In fact, the above inequity holds for all values except $n=\log_2e$

Still, when summing up each of these infinite sequences, the result is $1$ in both cases.

So in essence I am looking for a "native" (philosophical if you will) explanation of this.

Thank You.

barak manos
  • 43,109
  • Before the comment "different elements can sum up to the same value" is made, I would to emphasize that I'm aware of that. Yet, the issue here is these two infinite sequences, where each element except for the first one is larger than its "counterpart". – barak manos Aug 07 '14 at 21:46
  • 1
    So I could tell you $997 + 1 + 1 =333 + 333 + 333$ - I'm still not sure I understand the issue – Mathmo123 Aug 07 '14 at 21:51
  • 2
    The second series converges a bit faster than the first, but both ultimately yield the same value. – Lucian Aug 07 '14 at 21:52
  • @Mathmo123: Here I am talking about an infinite number of elements, all of which are larger than their counterparts by a distinct factor of $\ln(2^n)$. – barak manos Aug 07 '14 at 21:53
  • 2
    Let me give you a simpler example. $1 + \sum_{n=1}^\infty \frac 1{2^n} = 2 = 1.5 + \sum_{n=2}^\infty \frac 1{2^n}$. In this sequence you have the exact same issue. The point is that the difference in the sum of the remaining counterparts doesn't make up for the initial difference in the first term – Mathmo123 Aug 07 '14 at 21:55
  • What you are saying holds true for series with a finite number of terms. However, in this case, the number of terms is infinite. – Lucian Aug 07 '14 at 21:55
  • @ barak manos do you want a proof of the two identities or are you just concerned with their equality? – Asier Calbet Aug 07 '14 at 21:57
  • @Mathmo123: You'd have to agree that in my example, all the terms except for a finite amount (of $1$ term) are different, regardless of the order in which you arrange them. – barak manos Aug 07 '14 at 21:58
  • @Assaultous2: Well, the proof is obvious. They're both equal $1$. I'm wondering about the "miracle" here. – barak manos Aug 07 '14 at 21:59
  • no, proving that the second is equal to 1 is not obvious, but doable using a cool calculus trick – Asier Calbet Aug 07 '14 at 22:00
  • @Assaultous2: OK, I guess that basically you can show that the difference between the first term on each sequence is equal to the sum of the differences of the other terms on each sequence, right? – barak manos Aug 07 '14 at 22:01
  • nope. I'll show you in an answer if you want. You do know calculus, right? – Asier Calbet Aug 07 '14 at 22:03
  • @Assaultous2: I suppose so... – barak manos Aug 07 '14 at 22:03
  • Do you also consider $\sum_{n=1}^\infty1/2^n=\sum_{n=1}^\infty2/3^n$ a "miracle"? Sometimes an equality is just an equality. –  Aug 07 '14 at 22:47
  • 1
    The rigorous approach proves they're equal. However, it's not so difficult to reconcile this fact with a "naive" approach as you want to call it. $\ln(2^n)<1$ for $n=1$ but $\ln(2^n)>1$ for all other values. That's the key. The initial term is greater on the RHS (smaller denominator), but the other terms are all greater, so the sum "balances out" in the end. – Deepak Aug 08 '14 at 00:31

3 Answers3

10

Ok. So, lets define a function $f$ by the power series: $$ f(x)=\sum _{n=1}^{\infty} \frac{x^n}{n} $$ To find the closed form of this function, we first take its derivative: $$ f'(x) =\sum _{n=1}^{\infty} \frac{n x^{n-1}}{n} = \sum _{n=1}^{\infty} x^{n-1} = \frac {1}{1-x}$$ where the last equality is by the sum of an infinite geometric series and holds iff $|x|<1$. Integrating to recover $f$, we get: $$f(x)= - \ln(1-x) + C $$ and setting $x=0$, we see that $C=0$, so we have: $$f(x)= - \ln(1-x)$$ Letting $x=\frac{1}{2}$ and using both our power series and closed form of $f$, we have: $$ \sum _{n=1}^{\infty} \frac{1}{2^n n} = -\ln (1-\frac{1}{2})= \ln 2$$ Finally, $$ \sum _{n=1}^{\infty} \frac{1}{2^n \ln (2^n)} = \sum _{n=1}^{\infty} \frac{1}{2^n n \ln2} = \frac{1}{\ln 2} \sum _{n=1}^{\infty} \frac{1}{2^n n} = \frac{\ln 2}{\ln 2} =1 $$ Voila.

Asier Calbet
  • 2,480
5

It turns out that:

$$\sum_{n=1}^\infty \frac{1}{2^nn} = \ln 2$$

So

$$\sum_{n=1}^\infty \frac{1}{2^n\ln (2^n)} = \sum_{n=1}^\infty \frac{1}{2^nn\ln (2)} = \frac{1}{\ln 2}\sum_{n=1}^\infty \frac{1}{2^nn} = 1$$

Darth Geek
  • 12,296
2

$$A_n = \frac{1}{2^n}$$ $$B_n = \frac{1}{2^n\ln(2^n)}$$

You believe since $A_n > B_n$ that $\sum A_n > \sum B_n$ (assuming everything converges) and that is actually true. But there is one problem: $A_n > B_n$ is not true for $n=1$.

You could correctly reason that $$(\forall n > \color{red} 2: A_n > B_n) \Rightarrow \sum_{n=\color{red}2} A_n > \sum_{n=\color{red}2} B_n$$

but how much bigger is it? Well the difference comes from the $n=1$ term:

$$\sum_{n=2} A_n - \sum_{n=2} B_n = B_1 - A_1 = \frac{1}{2\ln2} - \frac 12 \approx 0.2213$$

DanielV
  • 23,556