Examine the convergence of the series: sum from n=1
Examine the convergence of the series: $$ \sum_{n=1}^{\infty} \frac{a+(-1)^n b}{a^n}, \qquad a>b>1. $$
Explanation of Method
Splitting the numerator into $a+(-1)^nb$ turns the series into the sum of two convergent geometric series — one plain, one alternating — each with ratio $1/a$ or $-1/a$ in absolute value less than $1$. Summing both closed forms gives the exact value, which is then cross-checked with the root test and a direct comparison bound.
- Split the series into two geometric series.
- Sum each part and combine.
- Verify using the root test and a comparison bound.
Step 1 · Split the series
$$ \sum_{n=1}^{\infty} \frac{a+(-1)^n b}{a^n} = a\sum_{n=1}^{\infty}\frac1{a^n} + b\sum_{n=1}^{\infty}\frac{(-1)^n}{a^n}. $$
Step 2 · Sum each part
Since $a>1$, both $\left|\frac1a\right|<1$ and $\left|-\frac1a\right|<1$, so both geometric series converge:
$$ a\sum_{n=1}^\infty\frac1{a^n} = a\cdot\frac{1/a}{1-1/a} = \frac{a}{a-1}, \qquad b\sum_{n=1}^\infty\left(\frac{-1}a\right)^n = b\cdot\frac{-1/a}{1+1/a} = \frac{-b}{a+1}. $$
Therefore the series converges, with sum $S=\dfrac{a}{a-1}-\dfrac{b}{a+1} = \dfrac{a^2+a(1-b)+b}{a^2-1}$.
Step 3 · Verify by root test and comparison
Root test: $\sqrt[n]{u_n}=\dfrac{\sqrt[n]{a+(-1)^nb}}{a}\to\dfrac1a<1$ (numerator, either $a-b$ or $a+b$ depending on parity, has $n$-th root $\to1$). Comparison: $|u_n|\le\dfrac{a+b}{a^n}$ for all $n$, and $\sum\dfrac{a+b}{a^n}$ is a convergent geometric series, so by comparison the original series converges absolutely.
Summary Table
| Quantity | Value |
|---|---|
| $\varlimsup\sqrt[n]{u_n}$ | $1/a<1$ |
| Sum $S$ | $\dfrac{a}{a-1}-\dfrac{b}{a+1}$ |
| Conclusion | Converges absolutely for $a>b>1$ |
Additional Notes
- The condition $a>b>1$ is only needed to guarantee $a>1$ (for the underlying geometric series to converge) — the constraint $a>b$ itself never actually enters the convergence argument, only the exact value of the sum.
- This construction — splitting a series with an oscillating numerator into a "plain" and an "alternating" geometric series — is a standard technique that generalizes to any numerator of the form $p+(-1)^nq$ divided by any base $>1$.
- Compare with Problem 82, where the oscillating factor is $\cos^2\frac{n\pi}3$ (period $6$) instead of $(-1)^n$ (period $2$): the same overall strategy of bounding the oscillating factor and comparing to a known convergent series applies, just with more residue classes to track.
Result
The series converges absolutely for $a>b>1$: