Find the sum sum from n=1 to infinity frac x^n n
Find the sum $\displaystyle\sum_{n=1}^{\infty}\frac{x^n}n$.
Explanation of Method
Integrating the geometric series $1/(1-t)=\sum t^n$ from $0$ to $x$ produces exactly this series on the right and $-\ln(1-x)$ on the left.
- Integrate the geometric series term by term from $0$ to $x$.
- Determine the exact region of convergence, including both endpoints.
- Verify numerically at a sample point.
Step 1 · Integrate the geometric series
$$ \frac1{1-t} = \sum_{n=0}^{\infty}t^n \implies \int_0^x\frac{dt}{1-t} = \sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1} = \sum_{n=1}^{\infty}\frac{x^n}n. $$ Since $\int_0^x\frac{dt}{1-t}=-\ln(1-x)$: $$ \sum_{n=1}^{\infty}\frac{x^n}n = -\ln(1-x). $$
Step 2 · Region of convergence
By the ratio test, $R=1$. At $x=1$, the series is the divergent harmonic series. At $x=-1$, it is the (convergent) alternating harmonic series, giving $-\ln2$. So the series converges on $-1\le x<1$.
Step 3 · Numerical check
At $x=\frac12$: $-\ln(1-\tfrac12)=\ln2\approx0.693147$; partial sums $0.5,0.625,0.6667,0.6823,\ldots$ approach this value. ✓
Summary Table
| Property | Value |
|---|---|
| Sum | $-\ln(1-x)$ |
| Region of validity | $-1\le x<1$ |
Additional Notes
- At $x=-1$ this becomes exactly Problem 289's alternating harmonic series, giving $-\ln2$ — the two problems are the same identity evaluated at opposite endpoints of convergence.
- Unlike the $x=-1$ endpoint (conditionally convergent), the $x=1$ endpoint diverges outright since $-\ln(1-x)\to\infty$ there, matching the divergence of the harmonic series.
- This is the basic building block behind Problem 261's factored-quadratic logarithm and Problem 272's $\ln\frac{1+x}{1-x}$, both constructed by combining shifted or signed copies of this same series.