Evaluate the infinite sum of 1/(n · 2ⁿ) for n from 1 to infinity.

Step-by-step solution

1. 1

   Check convergence before evaluating. The terms are positive and $\frac{1}{n2^{n}} \le \frac{1}{2^{n}}$, whose sum is the convergent geometric series $\sum 2^{-n} = 1$. So the series converges and its value is at most $1$ — a useful bound to compare the final answer against.

2. 2

   Recall the logarithmic (Mercator) series. For $|x| < 1$,

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

   This is the antiderivative of the geometric series $\frac{1}{1-x} = \sum_{n \ge 0} x^{n}$, integrated term by term from $0$ to $x$.

3. 3

   Rewrite the given term to match the pattern. Split the $2^{n}$ into the power of a single number:

   $\frac{1}{n\,2^{n}} = \frac{\left(\tfrac12\right)^{n}}{n}$

   so the series is exactly $\sum_{n\ge1} \frac{x^{n}}{n}$ with $x = \tfrac12$, comfortably inside the radius of convergence.

4. 4

   Substitute x = 1/2.

   $\sum_{n=1}^{\infty} \frac{\left(\tfrac12\right)^{n}}{n} = -\ln\left(1-\frac12\right) = -\ln\frac12$

5. 5

   Simplify the logarithm. Since $\ln\tfrac12 = -\ln 2$,

   $-\ln\frac12 = \ln 2$

   $\Longrightarrow \quad \sum_{n=1}^{\infty} \frac{1}{n\,2^{n}} = \ln 2$

6. 6

   Confirm numerically. Adding the first two hundred terms gives $0.6931471805599453$, and $\ln 2 = 0.6931471805599453$ — agreement to full double precision, and safely under the bound of $1$ from the first step.