Find lim as x to 1- sum from n=1 to infinity frac
Find $\displaystyle\lim_{x\to1^-}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n\frac{x^n}{1+x^n}$.
Explanation of Method
Rather than splitting the sum and separately estimating a tail (error-prone), the cleanest route is to show the series converges uniformly on $[0,1]$, which immediately makes the sum continuous at $x=1$ — so the limit equals the value of the series exactly at $x=1$.
- Show the series converges uniformly on $[0,1]$ via a uniform alternating-series remainder bound.
- Conclude continuity at $x=1$, so the limit equals the sum evaluated there.
- Evaluate the sum at $x=1$ using $\sum(-1)^{n-1}/n=\ln2$.
Step 1 · Uniform convergence on $[0,1]$
Let $a_n(x)=\dfrac1n\cdot\dfrac{x^n}{1+x^n}$ for $x\in[0,1]$. Since $0\le x^n/(1+x^n)<1$, and both $1/n$ and $x^n/(1+x^n)$ are non-negative and decreasing in $n$ (for fixed $x\in[0,1]$), the product $a_n(x)$ is decreasing in $n$ and $a_n(x)\to0$. The alternating-series remainder bound gives $$ |R_N(x)| \le a_{N+1}(x) = \frac1{N+1}\cdot\frac{x^{N+1}}{1+x^{N+1}} < \frac1{N+1} $$ for every $x\in[0,1]$ — an $x$-independent bound tending to $0$, so the series converges uniformly on $[0,1]$.
Step 2 · Continuity at $x=1$
Since each term $a_n(x)$ is continuous and the series converges uniformly on $[0,1]$, the sum $f(x)=\sum(-1)^{n-1}a_n(x)$ is continuous on $[0,1]$. In particular: $$ \lim_{x\to1^-}f(x) = f(1). $$
Step 3 · Evaluate $f(1)$
At $x=1$: $\dfrac{x^n}{1+x^n}=\dfrac{1}{2}$ for every $n$, so $$ f(1) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n\cdot\frac12 = \frac12\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}n = \frac12\ln2. $$
Summary Table
| Step | Result |
|---|---|
| Uniform bound | $|R_N(x)|<1/(N+1)$ on $[0,1]$ |
| $f(1)$ | $\tfrac12\ln2$ |
| Limit | $\tfrac12\ln2$ |
Additional Notes
- This is another instance of the technique used in Problem 201: a uniform alternating-series remainder bound rescues a series where the Weierstrass M-test would need more care, and here it does double duty by also delivering continuity right up to the boundary point $x=1$.
- The value $\ln2=\sum(-1)^{n-1}/n$ is the alternating harmonic series, the same constant that appears throughout this problem collection (e.g. Problems 141 and 155's evaluations involving $\ln2$).
- This problem is structurally an Abel's-theorem-style result: it justifies evaluating a limit "at the boundary of convergence" by exchanging a limit with an infinite sum, the same maneuver used (via a different mechanism — Abel summation) in Problem 205.