Using D'Alembert's ratio test, determine the convergence
Using D'Alembert's ratio test, determine the convergence of: $$ \sum_{n=1}^{\infty} \frac{2^n}{n}. $$
Explanation of Method
The ratio $\frac{a_{n+1}}{a_n}=2\cdot\frac n{n+1}\to2$: each term is asymptotically twice the previous one, reflecting the fact that the exponential factor $2^n$ dominates the merely linear denominator $n$. Since $2>1$, the ratio test gives divergence directly, and the Divergence Test independently confirms it since the terms grow without bound.
- Compute the ratio $a_{n+1}/a_n$ and its limit.
- Apply the ratio test to conclude divergence.
- Confirm independently via the Divergence Test.
Step 1 · Compute the ratio
With $a_n=\dfrac{2^n}n$:
$$ \frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{n+1}\cdot\frac{n}{2^n} = 2\cdot\frac{n}{n+1} \longrightarrow 2. $$
Step 2 · Apply the ratio test
Since $L=2>1$, by D'Alembert's ratio test $\sum_{n=1}^{\infty}\dfrac{2^n}n$ diverges.
Step 3 · Confirm via the divergence test
Since $2^n$ grows exponentially while $n$ grows only linearly, $\dfrac{2^n}n\to\infty\ne0$, independently confirming divergence.
Summary Table
| Quantity | Value |
|---|---|
| Ratio $a_{n+1}/a_n$ | $2\cdot\dfrac n{n+1}$ |
| $L=\lim a_{n+1}/a_n$ | $2$ |
| Conclusion | Diverges |
Additional Notes
- The terms grow explosively: $a_{10}=\frac{2^{10}}{10}=102.4$, $a_{20}=\frac{2^{20}}{20}\approx52{,}428.8$, confirming there is no hope of the tail sums settling down.
- This is the exact opposite scenario to Problem 58: there, factorial growth in the denominator crushed everything into rapid convergence; here, only linear growth ($n$) sits in the denominator, far too weak to counteract the numerator's exponential growth.
- This same ratio-test computation, generalized to $\sum a^n/n^y$ for arbitrary base $a$ and exponent $y$, is worked out fully in Problem 64 — this problem is the special case $a=2,\;y=1$.
Result
The series diverges: