Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=0}^{\infty} 2^n \sin\frac{1}{3^n}. $$
Explanation of Method
As $n\to\infty$, $\frac1{3^n}\to0$, so $\sin\frac1{3^n}\sim\frac1{3^n}$, making the general term behave like $2^n\cdot\frac1{3^n}=\left(\frac23\right)^n$ — a geometric series with ratio $\frac23<1$. Since exponential decay dominates over the exponentially growing factor $2^n$, the limit comparison test with this geometric series confirms convergence.
- Approximate $\sin\frac1{3^n}$ for large $n$ using its Taylor series.
- Identify the dominant behavior $2^n\sin\frac1{3^n}\sim\left(\frac23\right)^n$.
- Apply the limit comparison test against the convergent geometric series $\sum(2/3)^n$.
Step 1 · Taylor expand the sine term
$$ \sin\frac1{3^n} = \frac1{3^n} - \frac1{6\cdot3^{3n}} + O\!\left(\frac1{3^{5n}}\right). $$
Step 2 · Multiply by $2^n$
$$ 2^n\sin\frac1{3^n} = \left(\frac23\right)^n - \frac16\left(\frac2{27}\right)^n + O\!\left(\left(\frac2{243}\right)^n\right). $$
Every ratio appearing ($\frac23$, $\frac2{27}$, etc.) is less than $1$, so each individual piece is a convergent geometric series, and the dominant term is $\left(\frac23\right)^n$.
Step 3 · Limit comparison test
With $b_n=\left(\dfrac23\right)^n$, setting $h=\frac1{3^n}\to0^+$ and using $\lim_{h\to0}\frac{\sin h}h=1$:
$$ \lim_{n\to\infty}\frac{2^n\sin(1/3^n)}{(2/3)^n} = \lim_{n\to\infty} 3^n\sin\frac1{3^n} = 1. $$
Positive and finite; since $\sum\left(\frac23\right)^n$ is a convergent geometric series ($r=\frac23<1$), by the limit comparison test $\sum_{n=0}^{\infty}2^n\sin\frac1{3^n}$ converges too.
Summary Table
| Quantity | Value |
|---|---|
| General term asymptotics | $2^n\sin(1/3^n)\sim(2/3)^n$ |
| Geometric ratio | $\dfrac23<1$ |
| Limit comparison ratio | $1$ |
| Conclusion | Converges |
Additional Notes
- The dominant geometric series $\sum(2/3)^n$ sums to $\dfrac1{1-2/3}=3$, so the total sum of the original series is close to $3$, with small corrections from the higher-order terms in the sine expansion.
- The key balance is between the growth rate of $2^n$ and the decay rate of $\sin(1/3^n)\sim1/3^n$: since $2<3$, the product decays geometrically; if the bases were reversed (e.g. $3^n\sin(1/2^n)$), the growth would dominate and the series would diverge.
- Unlike the polynomial-rate series in earlier problems, convergence here is exponentially fast — the tail of the series shrinks by a constant factor of $\frac23$ with each additional term, so only a handful of terms are needed for high numerical precision.
Result
The series converges: