A sequence is defined by a_1=1
A sequence is defined by $a_1=1$ and $\dfrac{a_{n+1}}{a_n}=\dfrac34+\dfrac{(-1)^n}2$ for $n\ge1$. Test the absolute and conditional convergence of $\sum_{n=1}^{\infty}(-1)^n a_n$, and find its sum if it converges.
Explanation of Method
The ratio $a_{n+1}/a_n$ alternates between $\frac14$ (odd $n$) and $\frac54$ (even $n$), so consecutive pairs of terms combine into a single fixed ratio $\frac14\cdot\frac54=\frac5{16}$ every two steps. This lets the sequence be solved explicitly in closed form separately for odd and even indices, reducing the whole problem to two ordinary geometric series that can be summed directly — no test is needed once the closed form is found, only careful bookkeeping of the alternating sign.
- Derive closed-form expressions for $a_{2k-1}$ and $a_{2k}$.
- Show absolute convergence via the geometric ratio $5/16$.
- Sum the two resulting geometric series (odd- and even-indexed) to find the total.
Step 1 · Closed form via the two-step recursion
For odd $n=2k-1$: $\dfrac{a_{2k}}{a_{2k-1}}=\dfrac34+\dfrac{(-1)^{2k-1}}2=\dfrac34-\dfrac12=\dfrac14$. For even $n=2k$: $\dfrac{a_{2k+1}}{a_{2k}}=\dfrac34+\dfrac{(-1)^{2k}}2=\dfrac34+\dfrac12=\dfrac54$. So $\dfrac{a_{2k+1}}{a_{2k-1}}=\dfrac14\cdot\dfrac54=\dfrac5{16}$, a fixed ratio every two steps. With $a_1=1$: $a_{2k-1}=\left(\dfrac5{16}\right)^{k-1}$, and $a_{2k}=\dfrac14a_{2k-1}=\dfrac14\left(\dfrac5{16}\right)^{k-1}$.
Step 2 · Absolute convergence
Both $a_{2k-1}$ and $a_{2k}$ are (up to constant factors) geometric with ratio $\dfrac5{16}<1$ in $k$, so $a_n\to0$ geometrically and $\sum|a_n|=\sum a_n$ (all terms positive by construction) converges absolutely — hence $\sum(-1)^na_n$ converges absolutely too, by direct comparison.
Step 3 · Sum the series
Splitting by parity, with $r=\dfrac5{16}$: odd-indexed terms have sign $(-1)^{2k-1}=-1$, even-indexed terms have sign $(-1)^{2k}=+1$:
$$ \sum_{n=1}^{\infty}(-1)^na_n = -\sum_{k=1}^{\infty}a_{2k-1} + \sum_{k=1}^{\infty}a_{2k} = -\sum_{k=1}^{\infty}r^{k-1} + \frac14\sum_{k=1}^{\infty}r^{k-1} = \left(\frac14-1\right)\cdot\frac1{1-r}. $$
$$ = -\frac34\cdot\frac1{1-\frac5{16}} = -\frac34\cdot\frac{16}{11} = -\frac{12}{11}. $$
For reference, the absolute series sums to $\sum a_n=\left(1+\frac14\right)\cdot\dfrac1{1-\frac5{16}}=\dfrac54\cdot\dfrac{16}{11}=\dfrac{20}{11}$.
Summary Table
| Quantity | Value |
|---|---|
| Two-step ratio | $5/16$ |
| $\sum(-1)^na_n$ | $-12/11$ |
| $\sum a_n$ (absolute series) | $20/11$ |
Additional Notes
- Even though the ratio $a_{n+1}/a_n$ is $\frac54>1$ every other step, the sequence still decays overall because the geometric mean of $\frac14$ and $\frac54$ is $\sqrt{5}/4\approx0.559<1$; it is only the compounded two-step product $\frac5{16}$ that governs the true asymptotic decay rate, a subtlety the naive single-step ratio test would miss without splitting by parity.
- This recursive-sequence construction is a discrete analogue of Problems 131/135's parametrized alternating series: instead of an explicit formula, the terms are defined implicitly by a periodically-varying ratio, yet the resulting series analysis reduces to the same elementary geometric-series toolkit once the periodicity is unwound.
- This is one of the few problems in this range (alongside Problem 140) with a fully closed-form rational sum, making it a useful numeric checkpoint against the many "test only" problems (129–138) that establish convergence type without computing an explicit value.
Result
The series converges absolutely, with sum: