1^()circ Determine the interval of convergence
$1^\circ$ Determine the interval of convergence of $f_\lambda(x)=\sum_{n=1}^{\infty}n^\lambda x^{n-1}$ and examine the endpoints. $2^\circ$ Show $f_{\lambda+1}(x)=\dfrac d{dx}[xf_\lambda(x)]$.
Explanation of Method
The ratio test gives radius $1$ regardless of $\lambda$, since $(1+1/n)^\lambda\to1$; the endpoints then reduce to standard $p$-series and alternating-series tests. The differentiation identity follows immediately from term-by-term differentiation after multiplying by $x$.
- Apply the ratio test to find the radius.
- Test each endpoint separately using $p$-series and alternating-series criteria.
- Differentiate $xf_\lambda(x)$ term by term to establish the identity.
Part 1° · Radius and endpoints
$$ \left|\frac{a_{n+1}}{a_n}\right| = \left(1+\frac1n\right)^\lambda|x| \longrightarrow |x| \implies R=1 \text{ for every } \lambda. $$ At $x=1$: $f_\lambda(1)=\sum n^\lambda$ converges (as a $p$-series) iff $\lambda<-1$. At $x=-1$: $\sum(-1)^{n-1}n^\lambda$ has terms $\to0$ decreasingly precisely when $\lambda<0$, so by the Leibniz test it converges iff $\lambda<0$ (absolutely for $\lambda<-1$, conditionally for $-1\le\lambda<0$).
Part 2° · The differentiation identity
$$ xf_\lambda(x) = \sum_{n=1}^{\infty}n^\lambda x^n \implies \frac d{dx}[xf_\lambda(x)] = \sum_{n=1}^{\infty}n^{\lambda+1}x^{n-1} = f_{\lambda+1}(x). $$
Summary Table
| Property | Value |
|---|---|
| Radius | $R=1$, all $\lambda$ |
| At $x=1$ | converges iff $\lambda<-1$ |
| At $x=-1$ | converges iff $\lambda<0$ |
| Identity | $f_{\lambda+1}(x)=\frac d{dx}[xf_\lambda(x)]$ |
Additional Notes
- At $\lambda=0$: $f_0(x)=1/(1-x)$, and the identity recovers $f_1(x)=\frac d{dx}[x/(1-x)]=1/(1-x)^2$, matching the direct differentiation-based derivation of $\sum nx^{n-1}$ used in Problem 297.
- Applying the identity twice from $f_0$ reproduces exactly $f_2(x)=(1+x)/(1-x)^3$, Problem 297's result — this recursive machine generates $f_k$ for every non-negative integer $k$ purely by repeated differentiation and multiplication by $x$.
- For non-integer $\lambda$, $f_\lambda(x)$ is related to the polylogarithm: $xf_\lambda(x)=\operatorname{Li}_{-\lambda}(x)$, connecting this whole family to the special functions appearing in Problems 264 and 300.