Determine whether the series sum from n = 1 to infinity of n/3^n converges or diverges, and find its sum.

Step-by-step solution

1. 1

   Pick a test that suits the shape of the terms. The general term $a_n = n/3^n$ mixes a polynomial factor with an exponential one. Ratios of such terms simplify beautifully because $3^{n+1}/3^n$ collapses to a constant, so the ratio test is the natural choice — a comparison test would also work but needs a cleverly chosen comparison series.

2. 2

   Form the ratio of consecutive terms. With $a_n = \dfrac{n}{3^n}$ and $a_{n+1} = \dfrac{n+1}{3^{n+1}}$:

   $\frac{a_{n+1}}{a_n} = \frac{n+1}{3^{n+1}} \cdot \frac{3^n}{n} = \frac{n+1}{n}\cdot\frac{1}{3}$

   All terms are positive, so absolute values change nothing.

3. 3

   Take the limit and apply the test. Since $\dfrac{n+1}{n} = 1 + \dfrac{1}{n} \to 1$,

   $L = \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| = \frac{1}{3}$

   Because $L = \tfrac13 < 1$, the ratio test guarantees the series converges absolutely. Note the polynomial factor $n$ has no effect on the limit — geometric decay always beats polynomial growth.

4. 4

   Get the exact sum from the arithmetico-geometric formula. Convergence alone does not give a value, but this series is a standard one. Differentiating the geometric series $\sum_{n\ge0} r^n = \dfrac{1}{1-r}$ and multiplying by $r$ yields, for $|r|<1$,

   $\sum_{n=1}^{\infty} n r^{n} = \frac{r}{(1-r)^2}$

5. 5

   Substitute $r = 1/3$. Writing $\dfrac{n}{3^n} = n\left(\dfrac13\right)^{n}$:

   $\sum_{n=1}^{\infty}\frac{n}{3^n} = \frac{\tfrac13}{\left(1-\tfrac13\right)^2} = \frac{\tfrac13}{\tfrac49} = \frac13\cdot\frac94 = \frac34$

6. 6

   Sanity-check numerically. Adding the first few terms gives $0.3333 + 0.2222 + 0.1111 + 0.0494 + 0.0206 + \cdots$, and the partial sums climb to $0.75000$ and stay there — matching the exact value $3/4$.