Using Cauchy's root test, examine the convergence of: sum

Using Cauchy's root test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{n}{a^n}, \qquad a>1. $$

Step 1 · State the root test

For $b_n\ge0$, let $\rho=\limsup_{n\to\infty}\sqrt[n]{b_n}$: if $\rho<1$ the series converges, if $\rho>1$ it diverges, if $\rho=1$ the test is inconclusive.

Step 2 · Evaluate $\lim_{n\to\infty}\sqrt[n]n$

With $b_n=\dfrac n{a^n}$: $\sqrt[n]{b_n}=\dfrac{\sqrt[n]n}a$. Let $L=\lim_{n\to\infty}n^{1/n}$. Taking logs:

$$ \ln L = \lim_{n\to\infty}\frac{\ln n}n \overset{\text{L'Hôpital}}{=} \lim_{n\to\infty}\frac{1/n}1 = 0 \quad\Longrightarrow\quad L=e^0=1. $$

Step 3 · Combine and apply the root test

$$ \rho = \lim_{n\to\infty}\frac{\sqrt[n]n}a = \frac1a. $$

Since $a>1$, $\rho=\dfrac1a<1$, so by Cauchy's root test $\sum_{n=1}^{\infty}\dfrac n{a^n}$ converges.

Summary Table

Additional Notes

• The fact $\sqrt[n]n\to1$ is the same limit used throughout the earlier "$n$-th root" problems (e.g. Problems 16 and 17), confirming it as one of the handful of standard limits worth memorizing for root-test computations.
• This result matches what the ratio test would give: $\frac{a_{n+1}}{a_n}=\frac{n+1}{an}\to\frac1a$, the same limiting value $\rho$ — a useful consistency check, since ratio and root tests often (though not always) agree.
• The exact sum can also be found in closed form via differentiating the geometric series: $\sum_{n=1}^{\infty}\frac n{a^n} = \dfrac{1/a}{(1-1/a)^2} = \dfrac{a}{(a-1)^2}$, confirming the qualitative convergence result with an explicit value.