Limits Behind the Comparison Tests
The level 1 and level 2 limits you need for HW 5.4
The comparison tests come down to one question: what does \(a_n\) look like when \(n\) is huge? Once you know that \(a_n\) “behaves like” \(\frac{1}{n}\) or \(\frac{1}{n^2}\), the test is just paperwork. The limits on this page are the tools for answering that question.
Quick reminder of the two tests:
Suppose \(0 \le a_n \le b_n\) for all sufficiently large \(n\).
- If \(\sum b_n\) converges, then \(\sum a_n\) converges. (smaller than convergent ⇒ convergent)
- If \(\sum a_n\) diverges, then \(\sum b_n\) diverges. (bigger than divergent ⇒ divergent)
Suppose \(a_n, b_n > 0\) and \(\displaystyle L = \lim_{n\to\infty}\frac{a_n}{b_n}\).
- If \(0 < L < \infty\), then \(\sum a_n\) and \(\sum b_n\) both converge or both diverge.
- If \(L = 0\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
- If \(L = \infty\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
The Direct Comparison Test needs an inequality. The Limit Comparison Test needs a limit. Level 2 below gives you both kinds of fact.
Level 1: limits you should be able to do in your head
\(\displaystyle\lim_{n\to\infty}\frac{1}{n} = 0\)
As \(n\) grows, \(\frac1n\) shrinks toward \(0\): \(\frac1{10} = 0.1\), \(\frac1{1000} = 0.001\), \(\frac1{10^6} = 0.000001\). The same is true for \(\frac{1}{n^p}\) for any \(p > 0\), and for \(\frac{c}{n}\) for any constant \(c\).
This one matters more than it looks, because every substitution in Level 2 uses it. When we write \(t = \frac1n\), the statement “\(n \to \infty\)” turns into “\(t \to 0^+\).” That’s how limits at infinity become limits near zero, and back.
\(\displaystyle\lim_{n\to\infty}\frac{n}{n^2+n+1} = 0\)
Method: divide top and bottom by the highest power in the denominator. Here that’s \(n^2\):
\[ \frac{n}{n^2+n+1} = \frac{\frac{n}{n^2}}{\frac{n^2}{n^2}+\frac{n}{n^2}+\frac{1}{n^2}} = \frac{\frac1n}{1+\frac1n+\frac1{n^2}} \;\longrightarrow\; \frac{0}{1+0+0} = 0 \]
Dominant terms. For large \(n\), the \(n^2\) in the denominator swamps the \(n + 1\), so
\[ \frac{n}{n^2+n+1} \approx \frac{n}{n^2} = \frac1n \]
The Limit Comparison Test makes “\(\approx\)” precise. With \(b_n = \frac1n\):
\[ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{n}{n^2+n+1}\cdot\frac{n}{1} = \lim_{n\to\infty}\frac{n^2}{n^2+n+1} = 1 \]
\(0 < 1 < \infty\) and \(\sum\frac1n\) diverges, so \(\displaystyle\sum\frac{n}{n^2+n+1}\) diverges, even though its terms go to \(0\).
For a ratio of polynomials, keep only the highest power on top and the highest power on bottom. That’s your \(b_n\). Then the LCT limit is just the ratio of the leading coefficients.
Practice: \(\displaystyle\lim_{n\to\infty}\frac{3n^2-1}{5n^4+n}\), and use it to decide whether \(\displaystyle\sum\frac{3n^2-1}{5n^4+n}\) converges. Do the same for \(\displaystyle\sum\frac{1}{n-4}\) (compare it to \(\frac1n\); which one is bigger?).
Level 2: limits that take some work
These can’t be done by “plug in” or “divide by the highest power.” The main tools are:
- L’Hôpital’s rule for \(\frac00\) or \(\frac{\infty}{\infty}\) forms: \(\displaystyle\lim\frac{f}{g} = \lim\frac{f'}{g'}\).
- Take a log when the variable is in the exponent.
- Substitute \(t = \frac1n\) to move a limit at \(\infty\) to a limit near \(0\).
To use L’Hôpital on a sequence, replace \(n\) by a continuous variable \(x\) (derivatives need a function of a real variable). If \(f(x) \to L\) as \(x \to \infty\), then \(f(n) \to L\) too.
1. \(\displaystyle\lim_{n\to\infty} n^{1/n} = 1\)
This is an \(\infty^0\) form, and the answer isn’t obvious: the base grows while the exponent shrinks. When the variable is in the exponent, take a log.
Let \(y = x^{1/x}\). Then
\[ \ln y = \ln\left(x^{1/x}\right) = \frac{1}{x}\ln x = \frac{\ln x}{x} \]
That’s \(\frac{\infty}{\infty}\), so use L’Hôpital:
\[ \lim_{x\to\infty}\frac{\ln x}{x} = \lim_{x\to\infty}\frac{1/x}{1} = 0 \]
So \(\ln y \to 0\), which means \(y \to e^0 = 1\):
\[ \lim_{n\to\infty} n^{1/n} = 1 \]
| \(n\) | \(10\) | \(100\) | \(1000\) | \(10^6\) |
|---|---|---|---|---|
| \(n^{1/n}\) | \(1.2589\) | \(1.0471\) | \(1.0069\) | \(1.000014\) |
The limit you computed is the limit of \(\ln y\), which is \(0\). The limit of \(y\) itself is \(e^0 = 1\). Answering “\(0\)” is the most common mistake on this one.
2. \(\displaystyle\lim_{n\to\infty} \frac{n^{1+1/n}}{n} = 1\), so \(n^{1+1/n}\) behaves like \(n\)
\(n^{1+1/n}\) itself goes to \(\infty\); that part isn’t interesting. The useful question is how fast it grows. Split the exponent:
\[ n^{1+1/n} = n^1\cdot n^{1/n} \]
By limit 1, \(n^{1/n}\to 1\), so
\[ \lim_{n\to\infty}\frac{n^{1+1/n}}{n} = \lim_{n\to\infty} n^{1/n} = 1 \]
Application. \(\displaystyle\sum_{n=1}^\infty \frac{2}{n^{1+1/n}}\). With \(b_n = \frac1n\):
\[ L = \lim_{n\to\infty}\frac{2/n^{1+1/n}}{1/n} = \lim_{n\to\infty}\frac{2n}{n\cdot n^{1/n}} = \lim_{n\to\infty}\frac{2}{n^{1/n}} = \frac21 = 2 \]
\(0 < 2 < \infty\) and the harmonic series diverges, so this series diverges.
It’s tempting to say “\(p = 1 + \frac1n > 1\), so it converges.” But \(p\) has to be a constant. Here the exponent sinks toward \(1\), and the series behaves like the harmonic series.
3. \(\displaystyle\lim_{x\to 0}\ln(1+x) = 0\), and more usefully \(\displaystyle\lim_{x\to 0}\frac{\ln(1+x)}{x} = 1\)
The first limit is just continuity: \(\ln(1+x)\to\ln 1 = 0\). On its own that isn’t much help, because lots of things go to \(0\). What we want is how fast it goes to \(0\), and for that we compare with \(x\):
\[ \lim_{x\to0}\frac{\ln(1+x)}{x} \;\overset{\frac00}{=}\; \lim_{x\to0}\frac{\frac{1}{1+x}}{1} = \frac{1}{1+0} = 1 \]
So near zero, \(\ln(1+x)\approx x\). (Geometrically, \(y = x\) is the tangent line to \(y = \ln(1+x)\) at \(x = 0\).)
| \(x\) | \(0.1\) | \(0.01\) | \(0.001\) |
|---|---|---|---|
| \(\frac{\ln(1+x)}{x}\) | \(0.9531\) | \(0.99503\) | \(0.99950\) |
Application. \(\displaystyle\sum_{n=1}^\infty \ln\left(1 + \frac{3}{\sqrt n}\right)\). The inside of the log goes to \(0\), so this is \(\ln(1 + \text{small}) \approx \text{small} = \frac{3}{\sqrt n}\). Compare to \(b_n = \frac{1}{\sqrt n}\) (\(p = \frac12\)), and substitute \(t = \frac{3}{\sqrt n}\), so \(t\to0^+\) as \(n\to\infty\):
\[ L = \lim_{n\to\infty}\frac{\ln\left(1+\frac{3}{\sqrt n}\right)}{\frac{1}{\sqrt n}} = \lim_{n\to\infty} 3\cdot\frac{\ln\left(1+\frac{3}{\sqrt n}\right)}{\frac{3}{\sqrt n}} = 3\lim_{t\to0^+}\frac{\ln(1+t)}{t} = 3\cdot 1 = 3 \]
\(0 < 3 < \infty\) and \(\sum\frac{1}{\sqrt n}\) diverges (\(p = \frac12 \le 1\)), so the series diverges.
To use \(\frac{\ln(1+t)}{t}\to1\), the denominator has to match the inside of the log exactly. If the inside is \(\frac{3}{\sqrt n}\) but you’re dividing by \(\frac{1}{\sqrt n}\), multiply and divide by \(3\) to make them match.
There is also an inequality version, useful for direct comparison: \(\ln(1+x) < x\) for \(x > 0\). (The graph of \(\ln(1+x)\) is concave down, so it lies below its tangent line.)
In summary
- near zero, \(\ln(1+x)\approx x\)
- \(\ln(1+x) < x\) for \(x > 0\).
4. \(\sin x < x\) for \(x > 0\), and \(\displaystyle\lim_{x\to0}\frac{\sin x}{x} = 1\)
Plugging in \(x = \frac1n\) gives the hint in HW 5.4: \(\sin\left(\frac1n\right) < \frac1n\).
The inequality. Let \(g(x) = x - \sin x\). Then
\[ g(0) = 0, \qquad g'(x) = 1 - \cos x \ge 0 \]
So \(g\) starts at \(0\) and never decreases. (\(g'\) is zero only at isolated points, so \(g\) is actually strictly increasing.) Therefore \(g(x) > 0\) for \(x > 0\), which means
\[ \sin x < x \quad\text{for } x > 0 \]
The limit. \(\frac00\) form, so L’Hôpital:
\[ \lim_{x\to0}\frac{\sin x}{x} = \lim_{x\to0}\frac{\cos x}{1} = 1 \]
So near zero, \(\sin x\approx x\). That’s the same “tangent line” story as \(\ln(1+x)\).
| \(x\) | \(0.1\) | \(0.01\) | \(0.001\) |
|---|---|---|---|
| \(\frac{\sin x}{x}\) | \(0.998334\) | \(0.9999833\) | \(0.99999983\) |
Application (direct comparison). \(\displaystyle\sum_{n=1}^\infty\frac{\sin(1/n)}{\sqrt n}\).
First check the terms are positive: \(0 < \frac1n \le 1 < \pi\), so \(\sin\left(\frac1n\right) > 0\). Then use \(\sin\left(\frac1n\right) < \frac1n\):
\[ 0 < \frac{\sin(1/n)}{\sqrt n} < \frac{1/n}{\sqrt n} = \frac{1}{n^{3/2}} \]
\(\sum\frac{1}{n^{3/2}}\) converges (\(p = \frac32 > 1\)), so the smaller series converges.
Same series, by limit comparison. With \(b_n = \frac{1}{n^{3/2}}\) and \(t = \frac1n\):
\[ L = \lim_{n\to\infty}\frac{\sin(1/n)/\sqrt n}{1/n^{3/2}} = \lim_{n\to\infty}\frac{\sin(1/n)}{1/n} = \lim_{t\to0^+}\frac{\sin t}{t} = 1 \]
The conclusion is the same, by a different route.
5. \(\ln x < x^c\) for large \(x\) (any \(c > 0\))
Logs grow slower than every positive power, even tiny ones like \(x^{0.01}\). To prove it, show the ratio goes to \(0\). It’s \(\frac{\infty}{\infty}\), so use L’Hôpital:
\[ \lim_{x\to\infty}\frac{\ln x}{x^c} = \lim_{x\to\infty}\frac{\frac1x}{c\,x^{c-1}} = \lim_{x\to\infty}\frac{1}{c\,x^{c}} = 0 \qquad (c > 0) \]
Since the ratio \(\frac{\ln x}{x^c}\) goes to \(0\), it is eventually less than \(1\), so
\[ \ln x < x^c \quad\text{for all sufficiently large } x \]
| \(x\) | \(10\) | \(100\) | \(10^4\) | \(10^6\) |
|---|---|---|---|---|
| \(\frac{\ln x}{\sqrt x}\) | \(0.728\) | \(0.461\) | \(0.092\) | \(0.014\) |
For \(c = \frac12\) the inequality already holds at \(x = 1\). For \(c = 0.1\), \(\ln x\) stays bigger than \(x^{0.1}\) until about \(x \approx 3.4\times10^{15}\). The comparison tests only care about large \(n\), so that’s fine; just don’t check \(n = 10\) and conclude the inequality is false.
Application 1 (log on top, convergent). \(\displaystyle\sum_{n=1}^\infty\frac{\ln n}{n^2}\). The \(\ln n\) is a nuisance, so trade it for a small power, say \(c = \frac12\):
\[ \frac{\ln n}{n^2} < \frac{n^{1/2}}{n^2} = \frac{1}{n^{3/2}} \]
\(\sum\frac{1}{n^{3/2}}\) converges, so \(\sum\frac{\ln n}{n^2}\) converges. The trick is to pick \(c\) small enough that the result is still a convergent \(p\)-series. Any \(c < 1\) works here.
Application 2 (log on bottom, divergent). \(\displaystyle\sum_{n=2}^\infty\frac{1}{\ln n}\). Using \(c = 1\): \(\ln n < n\), so
\[ \frac{1}{\ln n} > \frac1n \]
That’s bigger than the divergent harmonic series, so it diverges.
Application 3 (logs that help). \(\displaystyle\sum_{n=2}^\infty\frac{1}{(n\ln n)^3}\). Here you don’t need the power bound at all: \(\ln n > 1\) once \(n \ge 3\), so
\[ \frac{1}{(n\ln n)^3} < \frac{1}{n^3} \]
which converges.
- \(\ln n\) in the numerator, aiming for convergence: replace \(\ln n\) with the bigger \(n^c\) (small \(c\)). The new series is bigger, and it’s a convergent \(p\)-series.
- \(\ln n\) in the denominator, aiming for divergence: \(\ln n < n^c\) flips to \(\frac{1}{\ln n} > \frac{1}{n^c}\). The original is bigger than a divergent \(p\)-series.
- \(\ln n\) in the denominator, aiming for convergence: \(\ln n > 1\) (for \(n \ge 3\)) makes the terms smaller than the same thing without the log.
Practice: \(\displaystyle\sum\frac{\ln n}{n^{1.5}}\) (which \(c\) works?), \(\;\displaystyle\sum\frac{\ln n}{n}\), \(\;\displaystyle\sum\frac{1}{\sqrt{n}\,\ln n}\)
Two more from HW 5.4
These follow the same “what does it look like near zero?” pattern.
\(1 - \cos x \approx \frac{x^2}{2}\) near zero. L’Hôpital twice:
\[ \lim_{x\to0}\frac{1-\cos x}{x^2} = \lim_{x\to0}\frac{\sin x}{2x} = \frac12\lim_{x\to0}\frac{\sin x}{x} = \frac12 \]
So \(1 - \cos\left(\frac1n\right)\approx\frac{1}{2n^2}\). Compare \(\sum\left(1-\cos\frac1n\right)\) to \(\sum\frac{1}{n^2}\): \(L = \frac12\), so it converges.
\(\arctan n - \frac\pi2 \approx -\frac1n\) for large \(n\). Write it as a quotient, then L’Hôpital:
\[ \lim_{x\to\infty}\frac{\arctan x - \frac\pi2}{\frac1x} \overset{\frac00}{=} \lim_{x\to\infty}\frac{\frac{1}{1+x^2}}{-\frac{1}{x^2}} = \lim_{x\to\infty}\frac{-x^2}{1+x^2} = -1 \]
So \(\frac1n\left(\arctan n - \frac\pi2\right)\approx -\frac{1}{n^2}\). Every term is negative (since \(\arctan n < \frac\pi2\)), so factor out \(-1\) and compare the positive series to \(\frac{1}{n^2}\): it converges.
Summary: what things look like
| For large \(n\) (or \(t = \frac1n \to 0\)) | behaves like | key limit |
|---|---|---|
| \(\dfrac{n}{n^2+n+1}\) | \(\dfrac1n\) | ratio \(\to 1\) |
| \(n^{1/n}\) | \(1\) | \(\ln y = \frac{\ln n}{n}\to0\) |
| \(n^{1+1/n}\) | \(n\) | \(n^{1/n}\to1\) |
| \(\ln(1+t)\) | \(t\) | \(\frac{\ln(1+t)}{t}\to1\) |
| \(\sin t\) | \(t\) (and \(\sin t < t\)) | \(\frac{\sin t}{t}\to1\) |
| \(1-\cos t\) | \(\frac{t^2}{2}\) | \(\frac{1-\cos t}{t^2}\to\frac12\) |
| \(\arctan n - \frac\pi2\) | \(-\frac1n\) | \(n\left(\arctan n-\frac\pi2\right)\to-1\) |
| \(\ln n\) | smaller than any \(n^c\) | \(\frac{\ln n}{n^c}\to0\) |
The workflow for limit comparison: decide what \(a_n\) behaves like and use that as \(b_n\). Compute \(\lim\frac{a_n}{b_n}\), using one of the limits above to finish it. If \(L\) is a positive finite number, \(\sum a_n\) does whatever \(\sum b_n\) does.