# Evaluating the series $\sum\limits_{k=1}^\infty \frac{1}{k2^k}$

I'm trying to find $\sum_{k=1}^\infty \frac{1}{k2^k}$. In fact I'm trying to find that $C$ for which $P(\{k\})=\frac{C}{k2^k}$ is a probability measure on $\mathbb{N}$ but presumably the intended solution goes by finding the value of the sum. I want to check if the following is a correct calculation / proof. If I write out the first several terms I don't see a way to make things better.

$$1/2 + 1/8 + 1/24 + 1/64 + ...$$

I thought about the trick used for weakly proving the geometric sum formula: Call the sum $s$, then

$$s = 1/2 + 1/8 + ...$$ $$4s = 2 + 1/2 + 1/8 + ...$$ so

$$3s = 2$$

so $s=2/3$.

• This is $\ln 2$.
– lulu
Jan 18, 2017 at 0:50
• The problem is that $4\times (1/24)\neq 1/8$, so your attempted cancellation breaks down. Instead, you need to look at the Taylor series for $\ln(1-x)$ and figure out what value to plug in for $x$. Jan 18, 2017 at 0:52
• See if these search results help: approach0.xyz/search/… Jan 18, 2017 at 1:21
• Possible duplicate of Compute the sum of the series $\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}$ Feb 5, 2018 at 13:37

Hint:

$$\ln(1-x)=-\sum_{k=1}^\infty \dfrac{x^k}k\enspace\text{for }\;\lvert x\rvert<1.$$

Hint: if $$f(z) = \sum_{n=1}^\infty \dfrac{z^n}{n}$$ differentiate $f$ term-by-term.

Taylor Series Centered at 1
$(0 < x \le2)$

$ln(x) =\sum_{n=1}^{\infty}(-1)^{(n+1)}{(x-1)^n\over n}= (x-1) - {(x-1)^2\over2} + {(x-1)^3\over3} - {(x-1)^4\over4} + ... + (-1)^{(n+1)}{(x-1)^n\over n} +....$

$\ln({1\over2}) = -{1\over2} - {1\over 8} - {1\over24} - {1\over64} - ...$

$\ln2^{-1} = -{1\over2} - {1\over 8} - {1\over24} - {1\over64} - ...$

$-\ln2 = -{1\over2} - {1\over 8} - {1\over24} - {1\over64} - ...$

$\ln 2 = {1\over2} + {1\over 8} + {1\over24} + {1\over64}+ ...$

• Not a complete answer? You underestimate what qualifies as complete. Jan 18, 2017 at 2:16