1^()circ Examine convergence of sum from n=1 to infinity
$1^\circ$ Examine convergence of $\sum_{n=1}^{\infty}\dfrac{(3n+2)^\alpha}{n(n+1)(n+2)}x^n$. $2^\circ$ For $\alpha=1$, $x=1$, find the sum.
Explanation of Method
Since $(3n+2)^\alpha$ and $n(n+1)(n+2)$ are both polynomial-type growth (any fixed power $\alpha$ raised on a linear factor still gives ratio $\to1$), the radius of convergence is $1$ regardless of $\alpha$; endpoint behavior then follows from comparing $n^{\alpha-3}$ against $p$-series thresholds. For the concrete sum, partial fractions turn the series into a telescoping combination of harmonic numbers.
- Apply the ratio test to find $R=1$ for every $\alpha$.
- Compare against $p$-series at each endpoint.
- Decompose via partial fractions and sum by careful telescoping for $\alpha=1$, $x=1$.
Part 1° · Convergence
Since $\dfrac{(3n+5)^\alpha}{(3n+2)^\alpha}\to1$ and $\dfrac n{n+3}\to1$ regardless of $\alpha$, the ratio test gives $R=1$ for every $\alpha\in\mathbb R$. For large $n$, the general term behaves like $3^\alpha n^{\alpha-3}$: $$ \text{At } x=1:\ \text{converges} \iff \alpha-3<-1 \iff \alpha<2. $$ $$ \text{At } x=-1:\ \text{terms} \sim(-1)^n\cdot3^\alpha n^{\alpha-3}\to0 \iff \alpha<3, \text{ so the alternating series converges} \iff \alpha<3 \text{ (strictly — at }\alpha=3\text{ the terms tend to the nonzero constant }27\text{, so the series diverges).} $$ The series converges uniformly on any $[-r,r]$ with $r<1$, for every $\alpha$ (Weierstrass $M$-test).
Part 2° · Sum for $\alpha=1$, $x=1$
Partial fractions give (matching at $n=1$: $\tfrac56=1-\tfrac12+\tfrac23$... verified directly): $$ \frac{3n+2}{n(n+1)(n+2)} = \frac1n+\frac1{n+1}-\frac2{n+2}. $$ Splitting the summand as $\left(\dfrac1n-\dfrac1{n+2}\right)+\left(\dfrac1{n+1}-\dfrac1{n+2}\right)$, each piece telescopes separately: $$ \sum_{n=1}^N\left(\frac1n-\frac1{n+2}\right) = 1+\frac12-\frac1{N+1}-\frac1{N+2}, \qquad \sum_{n=1}^N\left(\frac1{n+1}-\frac1{n+2}\right) = \frac12-\frac1{N+2}. $$ Adding: $$ S_N = 2 - \frac1{N+1} - \frac2{N+2} \ \longrightarrow\ 2 \quad\text{as } N\to\infty. $$ (Numerical check: summing the first 10 terms directly gives $\approx1.742$, with the slowly-decaying $\sim3/n^2$ tail adding roughly $0.27$ more — consistent with a limit of $2$.)
Summary Table
| Property | Value |
|---|---|
| Radius | $R=1$, all $\alpha$ |
| At $x=1$ | converges iff $\alpha<2$ |
| At $x=-1$ | converges iff $\alpha<3$ |
| Sum ($\alpha=1,x=1$) | $2$ |
Additional Notes
- Splitting the telescoping sum into two separate "skip-one" and "skip-two" telescoping pieces (rather than trying to cancel all three partial-fraction terms directly) is the safest way to avoid sign errors — direct numerical verification (partial sums approaching $2$, not $1.5$) is a valuable independent check on any such computation.
- The boundary case $\alpha=3$ at $x=-1$ is a useful reminder that the alternating series test requires terms to actually tend to $0$: even though the series alternates, a nonzero limiting term (here $27$) makes it diverge outright, not merely "borderline."
- This partial-fraction-then-telescope technique is the same one used for the harmonic-type sums in Problems 290 and 296, all reducing rational-function-weighted series to combinations of telescoping differences.