Practice: Convergence of Infinite Series

(a) $\sum _{k=0}^{\mathrm{\infty }}{\displaystyle \frac{2}{5^k}}$. This is geometric with first term $a=2$ (at $k=0$) and ratio $r=\frac{1}{5}$. The licensing condition is , and ✓. It is what makes $r^{n+1}\to 0$ in the partial-sum formula $s_n=\frac{a(1-r^{n+1})}{1-r}$, and without it the series diverges.

Check: partial sums $2$, $2.4$, $2.48$, $2.496$ climb toward $2.5$ ✓. (b) $\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{4^n}{n!}}$. The factorial is the signature for the ratio test:

Since , the series converges absolutely. The factorial eventually outgrows any fixed power, which is the same comparison L'Hôpital established for $x^n/e^x$. (The sum is in fact $e^4-1$, though the test does not produce it.) (c) $\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{\sqrt{n}}}$. Why vanishing terms do not settle it. $a_n=n^{-1/2}\to 0$, so the $n$th-term test is silent. That test runs one way only: non-vanishing terms prove divergence, vanishing terms prove nothing. The harmonic series is the standing counterexample, terms vanish, partial sums grow like $\ln n$ past every bound. A test that decides. This is a $p$-series with $p=\frac{1}{2}$, and $p$-series converge exactly when . Since $\frac{1}{2}\le 1$, the series diverges. Equivalently by the integral test: $f(x)=x^{-1/2}$ is positive and decreasing on $[1,\mathrm{\infty })$, and

so the series diverges with it. Note this is the $p$-test threshold at infinity from the integration unit, read on a series instead of an integral.

(d) Absolute against conditional convergence. Consider $\sum (-1{)}^{n+1}/n$ alongside the series in (c). The alternating harmonic series converges by the alternating series test, since $1/n$ decreases to $0$, and its sum is $\ln 2\approx 0.693147$. But the series of magnitudes is the harmonic series, which diverges. The convergence is therefore conditional, and it rests entirely on cancellation between successive terms.

That matters because a conditionally convergent series may not be rearranged: reordering its terms can produce any real number, or divergence. Only absolute convergence, where $\sum |a_n|$ itself converges, makes the sum independent of the order of addition and permits the usual algebraic manipulations.