Practice Exam 3
Sequences & Infinite Series
On series questions, state which test you are applying, verify its hypotheses, and state the conclusion. A correct verdict with no named test earns partial credit at best — because the reasoning is the thing being assessed, not the yes/no answer.
Find \(\lim_{n\to\infty} a_n\) for \(a_n = \dfrac{\ln n}{\sqrt{n}}\), or show it does not exist.
Solution
Both parts \(\to\infty\), so this is \(\infty/\infty\). Replace \(n\) by a continuous \(x\) and apply L’Hôpital:
Or by growth ordering, with no work: \(\ln n\) is slower than every positive power of \(n\), including \(n^{1/2}\). The denominator wins, so the limit is 0.
Evaluate \(\displaystyle\sum_{n=1}^{\infty} 5\left(-\frac{1}{3}\right)^{n}\), or show it diverges.
Solution
Identify \(r\) and the first term. The ratio is \(r = -1/3\), and \(|r| = 1/3 < 1\), so it converges.
The sum starts at \(n = 1\), so the first term present is
The trap: reflexively writing \(a = 5\). That is the \(n = 0\) term, which is not in this sum, and would give 15/4 instead. Always evaluate the first index actually appearing.
Evaluate \(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n+1)(n+2)}\).
Solution
Partial fractions (Unit 1) on the general term:
Write out the partial sum — this is the step that must appear in your work:
Everything cancels except the first and last pieces:
Telescoping is one of only two families you can sum exactly (the other being geometric). Everywhere else you can decide convergence but not evaluate.
Determine whether \(\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{3n^2 + 1}\) converges.
Solution
Always check the terms first — it costs nothing:
By the \(n\)th-term test for divergence, the series diverges.
Intuition: you are adding numbers that stay near 1/3 forever. After \(N\) terms the sum is roughly \(N/3\), which grows without bound. No further test is needed, and applying one would waste exam minutes.
Determine whether \(\displaystyle\sum_{n=2}^{\infty}\frac{1}{n\ln n}\) converges.
Solution
Terms \(\to 0\), so the \(n\)th-term test is inconclusive. The ratio test also returns 1 here. Use the integral test with \(f(x) = 1/(x\ln x)\).
Check the hypotheses on \([2,\infty)\): \(f\) is continuous, positive, and decreasing (both factors in the denominator increase). ✓
The integral diverges, so the series diverges.
Worth appreciating: this diverges even though its terms shrink faster than the harmonic series’. The extra \(\ln n\) is simply too weak to rescue it. Yet \(\sum 1/\big(n(\ln n)^2\big)\) does converge — the same substitution gives \(\int du/u^2\), which is finite. The boundary between convergence and divergence is far finer than it first appears.
Determine whether \(\displaystyle\sum_{n=1}^{\infty}\frac{2n + 1}{n^3 + 5}\) converges.
Solution
Dominant behavior: \(2n/n^3 = 2/n^2\). Compare with \(b_n = 1/n^2\).
Finite and nonzero, so the series behave alike. \(\sum 1/n^2\) is a p-series with \(p = 2 > 1\) and converges, so the given series converges by limit comparison.
Shortcut for any rational series: subtract the numerator’s degree from the denominator’s. Here \(3 - 1 = 2\), so it behaves like a p-series with \(p = 2\). Convergent whenever that difference exceeds 1.
Determine whether \(\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{2^n}\) converges.
Solution
An exponential in the denominator points at the ratio test:
The series converges absolutely.
The general principle: exponential growth beats polynomial growth, so the \(n^2\) in the numerator never catches the \(2^n\). Any \(\sum n^k/c^n\) with \(c > 1\) converges, whatever the fixed power \(k\).
Determine whether \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{2n + 3}{3n + 2}\right)^{n}\) converges.
Solution
The entire term is raised to the \(n\)th power — that is the signature of the root test:
The series converges absolutely.
Why the root test and not the ratio test: the \(n\)th root annihilates the outer power in one step, leaving a trivial limit. The ratio test would produce a ratio of two \(n\)th powers with different bases — doable, but far messier. Matching the test to the structure is most of the skill.
For \(\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}\): (a) does it converge? (b) absolutely or conditionally? (c) how many terms guarantee an error under 0.01?
Solution
(a) Convergence. Apply the alternating series test to \(b_n = 1/\sqrt{n}\):
- \(b_n > 0\) ✓
- decreasing, since \(\sqrt{n}\) increases ✓
- \(\lim b_n = 0\) ✓
All three hold, so the series converges.
(b) Absolutely? Take absolute values:
p-series with \(p = 1/2 \le 1\), which diverges. Therefore the original is conditionally convergent.
(c) Error bound. The remainder is at most the first omitted term:
So \(N = 10\,000\) terms — for two decimal places.
The practical lesson: convergence and usable convergence are different things. Compare Exam-3 problem 4 of the unit page, where \(1/n^3\) needed only 10 terms for three decimals. A conditionally convergent series is often too slow to compute with directly, which is why numerical work goes to great lengths to find faster-converging equivalents.
A first-order digital filter obeys \(y[n] = x[n] + a\,y[n-1]\), giving impulse response \(h[n] = a^n\) for \(n \ge 0\).
(a) For \(a = 0.8\), find the DC gain \(\sum h[n]\). (b) Find the total energy \(\sum h[n]^2\). (c) What happens when \(a = 1.2\), and what is the general stability condition?
Solution
(a) DC gain. Geometric with \(a_{\text{first}} = 1\) (the \(n = 0\) term) and ratio 0.8:
A constant input of 1 eventually produces an output of 5 — a gain of 5 at DC.
(b) Energy. Squaring gives another geometric series, ratio \(a^2\):
Note this is smaller than the DC gain. Energy sums squares, and squaring numbers below 1 shrinks them — the same reason RMS and average differ in Unit 2.
(c) \(a = 1.2\). Now \(|r| = 1.2 > 1\), so
The impulse response grows without bound: 1, 1.2, 1.44, 1.73, … A single input sample makes the output increase forever. The filter is unstable.
This is the unit circle criterion. In the z-domain, \(a\) is the filter’s pole, and “all poles strictly inside the unit circle” is exactly the statement that the geometric series \(\sum a^n\) converges. You now know why that rule is true rather than merely what it says.
Edge case worth naming: \(a = 1\) gives \(h[n] = 1\) forever — the series diverges, but the output does not blow up; it accumulates. That is a perfect integrator, marginally stable. Useful on purpose, dangerous by accident.
Scoring
| Score | Where you are |
|---|---|
| 90–100 | Test selection is automatic. Move to power series. |
| 75–89 | Right tests, sloppy hypothesis checks. Write out the conditions explicitly. |
| 60–74 | You are reaching for tests semi-randomly. Drill Which test to use on twenty series without solving any of them. |
| < 60 | Re-read the unit. Series is the foundation of Unit 4 — a shaky base there gets expensive fast. |
Using the wrong first term in a geometric series (Q2); claiming convergence from \(a_n \to 0\); forgetting to check whether an alternating series converges absolutely before calling it conditional (Q9); and not verifying that \(f\) is decreasing before invoking the integral test (Q5). The last one is a hypothesis, not a formality — the test is simply invalid without it.