Both Term Tests Return One, and the Sum Still Stops Below Two

Add up $e^{-\sqrt{n}}$ for every whole number $n \ge 1$. The terms shrink, but slowly: the hundredth is $e^{-10} \approx 4.5 \times 10^{-5}$ and the ten-thousandth is still $e^{-100}$, which is small but arrived at unhurriedly. Does the total stay finite?

It does, and it lands at 1.6704068. The interesting part is that the two tests most people reach for first both refuse to answer, and that the standard write-up of this problem gets the bound backwards.

Two tests, no verdict

The ratio test looks at consecutive terms:

The exponent goes to zero like $1/(2\sqrt{n})$, so the ratio approaches 1 from below and never gets there. The root test does no better: the $n$-th root of the $n$-th term is $e^{-1/\sqrt{n}}$, which also tends to 1. A limit of 1 is the one value at which both tests are silent by construction, and silence is not divergence.

Nor does the shrinking of the terms settle anything. The harmonic series has terms going to zero and diverges anyway, so $a_n \to 0$ is a requirement rather than an argument. At this point the instruments have run out, and the move is to stop weighing individual terms and start measuring area instead.

Standing the terms up as bars

Consider the function $f(x) = e^{-\sqrt{x}}$ on the positive axis. It is decreasing everywhere, which the derivative confirms:

Now draw a bar for each term, one unit wide, with height $f(n) = e^{-\sqrt{n}}$. Its area is the term itself, so the sum of the series is the total area of the bars. The question is where to put each bar, and this is the step that decides everything.

Put the bar for term $n$ on the interval $[n-1, n]$. Its height is the value of $f$ at the right end of that interval, and since $f$ is decreasing, the right end is where $f$ is smallest there. So the bar sits entirely under the curve, every bar does, and the bars do not overlap:

Put the same bar on $[n, n+1]$ instead and the height becomes the value at the left end, which is the largest value on that interval, so each bar now covers the curve rather than hiding under it. That comparison runs the other way:

This is where the usual presentation of the problem goes wrong. It computes the integral from 1, gets $4/e$, and reports that the sum is just under that. The sum is not under $4/e$. It is 13.52 percent above it, because the integral from 1 is a floor. The cap you want is the integral from 0, and it happens to be a rounder number.

Evaluating both integrals

Substitute $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u\,du$. The improper integral becomes an ordinary one:

One integration by parts gives $\int u e^{-u} du = -(u+1)e^{-u}$, and the same antiderivative evaluated from $u = 1$ supplies the other integral, $2 \cdot 2/e = 4/e$. It also gives the tail in closed form, which is what makes a numerical check honest:

At $N = 4{,}000{,}000$ that tail is below $10^{-300}$, so the partial sum to four million terms, 1.6704068180, brackets the true value to a width no arithmetic will ever notice. Sharpening (4) by adding the first term to it gives the classical sandwich $4/e \le S \le e^{-1} + 4/e$, that is $1.4715 \le S \le 1.8394$, which is tighter than the cap of 2 and less quotable.

The same argument on a whole family

Nothing in the derivation used the square root beyond its monotonicity, so replace it with any positive power. For $\alpha > 0$, the substitution $u = x^{\alpha}$ turns the integral into a gamma function:

So $\sum e^{-n^{\alpha}}$ converges for every positive $\alpha$, with cap $\Gamma(1 + 1/\alpha)$. At $\alpha = 1/2$ that is $\Gamma(3) = 2$, which is where the 2 in (3) comes from. Both term tests still return 1 for every $\alpha < 1$, so the whole family is invisible to them and visible to one integral.

Where the method needs care

The monotonicity in (2) is not decoration, and it is easy to see what it is holding up. Take $g(x) = e^{-\sqrt{x}}\cos^{20}(\pi x)$. Since $\cos(\pi n) = \pm 1$, this agrees with our terms at every single integer, so the sum of $g(n)$ is still 1.6704068. But $g$ dips almost to zero between the integers, and its integral from 0 to infinity is about 0.366, well below the sum. A bound read off that curve would be flatly false. The sampled values do not control the area on their own; a decreasing curve is what ties the two together, because it attains its extremes at the ends of each interval.

The cap from 0 also requires the integrand to behave at the origin. Here $f(0) = 1$ and there is nothing to worry about, but for a function with a singularity at 0, say $f(x) = e^{-\sqrt{x}}/x$, the integral from 0 diverges and the only usable bound is $f(1) + \int_1^{\infty} f$. The version of the integral test that survives that case is the sandwich, not the clean cap.

One alternative is often suggested and worth pricing honestly. Comparing with $1/n^2$ does work, because $e^{-\sqrt{n}} \le 1/n^2$ exactly when $\sqrt{n} \ge 2\ln n$, and the difference $\sqrt{n} - 2\ln n$ is increasing past $n = 16$ and first turns positive at $n = 75$. So the comparison settles convergence, and it says nothing at all about the first seventy-four terms. You have to add those up by hand before the argument produces a number, which is why the integral is the route worth remembering.

Sources and further reading

• The comparison that decided the question — Integral test for convergence
• What a limit of 1 means, and does not mean — Ratio test and Root test
• The function that appears in (5) and (7) — Gamma function
• The object being evaluated in (3) and (4) — Improper integral

Both bar placements were checked term by term over the first two hundred indices and against fine Riemann sums of both integrals, the two term-test limits were evaluated out to $n = 10^{12}$, and the same machinery was run on the harmonic series as a control, where the integral diverges and the partial sums track $\ln n$ as they should.