TITLE: How come $\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}$?
QUESTION [6 upvotes]: 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.

REPLY [2 votes]: $$A_n = \frac{1}{2^n}$$
$$B_n = \frac{1}{2^n\ln(2^n)}$$
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:
$$\sum_{n=2} A_n - \sum_{n=2} B_n = B_1 - A_1 = \frac{1}{2\ln2} - \frac 12 \approx 0.2213$$