Determine the radius of convergence and test convergence

Determine the radius of convergence and test convergence at the endpoints for $\displaystyle\sum_{n=0}^{\infty}(n+1)x^n$.

Step 1 · Radius of convergence

$$ \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n\to\infty}\frac{n+2}{n+1} = 1 \implies R=1. $$

Step 2 · Endpoint $x=1$

$\sum_{n=0}^{\infty}(n+1)=1+2+3+\cdots$; since $(n+1)\to\infty\ne0$, the series diverges by the Divergence Test.

Step 3 · Endpoint $x=-1$

$\sum_{n=0}^{\infty}(-1)^n(n+1)=1-2+3-4+\cdots$; since $|(-1)^n(n+1)|=n+1\to\infty$, the terms do not tend to $0$, so the series diverges by the Divergence Test.

Summary Table

Additional Notes

• This series is exactly the term-by-term derivative of the geometric series (as computed in Problem 211): $\sum(n+1)x^n=\sum nx^{n-1}=1/(1-x)^2$, blowing up as $x\to1^-$, which is consistent with divergence at $x=1$.
• Since differentiating a power series never changes its radius of convergence, this radius of $1$ matches the parent geometric series $\sum x^n$ exactly — differentiation can only affect endpoint behavior, never the radius itself.
• This is the simplest possible example in the "diverges at both endpoints" category, contrasting with Problem 225's $\sum x^n/n$ (converges at one endpoint) and Problem 226's $\sum x^n/(n(n+1))$ (converges at both).