Practice Exam 3

Sequences & Infinite Series

Questions10

Time90 min

Points100

RuleName the test you use

Grading convention

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.

1Sequence limit8 pts

Find lim_n→∞ a_n for a_n = ln n√n, or show it does not exist.

Solution

Both parts → ∞, so this is ∞/∞. Replace n by a continuous x and apply L’Hôpital:

lim_x→∞ ln xx^1/2 = lim_x→∞ 1/x¹⁄₂x^−1/2 = lim_x→∞ 2x^1/2 = 0

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.

2Geometric series10 pts

Evaluate ∞Σn=1 5(−13)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

a = 5(−13)¹ = −53

S = a1 − r = −5/31 + 1/3 = −5/34/3 = −54

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.

3Telescoping10 pts

Evaluate ∞Σn=1 1(n+1)(n+2).

Solution

Partial fractions (Unit 1) on the general term:

1(n+1)(n+2) = 1n+1 − 1n+2

Write out the partial sum — this is the step that must appear in your work:

s_N = (12−13) + (13−14) + … + (1N+1−1N+2)

Everything cancels except the first and last pieces:

s_N = 12 − 1N+2 → 12 as N → ∞

Telescoping is one of only two families you can sum exactly (the other being geometric). Everywhere else you can decide convergence but not evaluate.

4nth-term test8 pts

Determine whether ∞Σn=1 n²3n² + 1 converges.

Solution

Always check the terms first — it costs nothing:

lim_n→∞ n²3n²+1 = 13 ≠ 0

By the nth-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.

5Integral test10 pts

Determine whether ∞Σn=2 1n ln n converges.

Solution

Terms → 0, so the nth-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,∞): f is continuous, positive, and decreasing (both factors in the denominator increase). ✓

∫₂^∞ dxx ln x substitute u = ln x, du = dx/x

= ∫_{ln 2}^∞ duu = lim_T→∞ [ln u]_{ln 2}^T = ∞

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 Σ1/(n(ln n)²) does converge — the same substitution gives ∫du/u², which is finite. The boundary between convergence and divergence is far finer than it first appears.

6Limit comparison10 pts

Determine whether ∞Σn=1 2n + 1n³ + 5 converges.

Solution

Dominant behavior: 2n/n³ = 2/n². Compare with b_n = 1/n².

lim_n→∞ (2n+1)/(n³+5)1/n² = lim_n→∞ 2n³ + n²n³ + 5 = 2

Finite and nonzero, so the series behave alike. Σ1/n² 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.

7Ratio test10 pts

Determine whether ∞Σn=1 n²2ⁿ converges.

Solution

An exponential in the denominator points at the ratio test:

a_n+1a_n = (n+1)²2ⁿ⁺¹ · 2ⁿn² = 12(n+1n)²

L = lim_n→∞ 12(1 + 1n)² = 12 < 1

The series converges absolutely.

The general principle: exponential growth beats polynomial growth, so the n² in the numerator never catches the 2ⁿ. Any Σn^k/cⁿ with c > 1 converges, whatever the fixed power k.

8Root test10 pts

Determine whether ∞Σn=1 (2n + 33n + 2)ⁿ converges.

Solution

The entire term is raised to the nth power — that is the signature of the root test:

L = lim_n→∞ ⁿ√(2n+33n+2)ⁿ = lim_n→∞ 2n+33n+2 = 23 < 1

The series converges absolutely.

Why the root test and not the ratio test: the nth root annihilates the outer power in one step, leaving a trivial limit. The ratio test would produce a ratio of two nth powers with different bases — doable, but far messier. Matching the test to the structure is most of the skill.

9Alternating · conditional12 pts

For ∞Σn=1 (−1)ⁿ√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/√n:

b_n > 0 ✓
decreasing, since √n increases ✓
lim b_n = 0 ✓

All three hold, so the series converges.

(b) Absolutely? Take absolute values:

∞Σn=1 1√n = ∞Σn=1 1n^1/2

p-series with p = 1/2 ≤ 1, which diverges. Therefore the original is conditionally convergent.

(c) Error bound. The remainder is at most the first omitted term:

1√(N+1) < 0.01 ⇒ √(N+1) > 100 ⇒ N + 1 > 10 000

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/n³ 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.

10Geometric · EE12 pts

A first-order digital filter obeys y[n] = x[n] + a y[n−1], giving impulse response h[n] = aⁿ for n ≥ 0.

(a) For a = 0.8, find the DC gain Σh[n]. (b) Find the total energy Σh[n]². (c) What happens when a = 1.2, and what is the general stability condition?

Solution

(a) DC gain. Geometric with a_first = 1 (the n = 0 term) and ratio 0.8:

∞Σn=0 (0.8)ⁿ = 11 − 0.8 = 5

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 a²:

∞Σn=0 (0.8ⁿ)² = ∞Σn=0 (0.64)ⁿ = 11 − 0.64 = 10.36 ≈ 2.78

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

∞Σn=0 (1.2)ⁿ diverges

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.

Stability condition: |a| < 1

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 Σaⁿ 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.

Where the points usually go

Using the wrong first term in a geometric series (Q2); claiming convergence from a_n → 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.