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

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.

#### Solutions Collecting From Web of "How come $\sum\limits_{n=1}^{\infty}\frac{1}{2^n}=\sum\limits_{n=1}^{\infty}\frac{1}{2^n\ln(2^n)}$?"

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.

It turns out that:

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

So

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

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

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:

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