The nth Term Test for Divergence - AP Calculus BC
Card 1 of 30
Evaluate $\lim_{{n \to \infty}} \frac{3n}{n+3}$. Divergent?
Evaluate $\lim_{{n \to \infty}} \frac{3n}{n+3}$. Divergent?
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Limit is 3; series diverges. Leading terms give ratio $\frac{3}{1} = 3 \neq 0$.
Limit is 3; series diverges. Leading terms give ratio $\frac{3}{1} = 3 \neq 0$.
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Evaluate $\lim_{{n \to \infty}} \frac{2}{n}$. Divergent?
Evaluate $\lim_{{n \to \infty}} \frac{2}{n}$. Divergent?
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Limit is 0; test inconclusive. Constant multiple of $\frac{1}{n}$ still gives limit 0.
Limit is 0; test inconclusive. Constant multiple of $\frac{1}{n}$ still gives limit 0.
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Assess $\lim_{{n \to \infty}} \frac{2n+1}{n}$. Divergent?
Assess $\lim_{{n \to \infty}} \frac{2n+1}{n}$. Divergent?
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Limit is 2; series diverges. Dividing gives $2 + \frac{1}{n}$, limit 2.
Limit is 2; series diverges. Dividing gives $2 + \frac{1}{n}$, limit 2.
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What if $\lim_{{n \to \infty}} a_n \neq 0$?
What if $\lim_{{n \to \infty}} a_n \neq 0$?
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Series $\sum a_n$ diverges. Non-zero limit always implies divergence by the test.
Series $\sum a_n$ diverges. Non-zero limit always implies divergence by the test.
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Find $\lim_{{n \to \infty}} \frac{n^2}{n^3 + 1}$.
Find $\lim_{{n \to \infty}} \frac{n^2}{n^3 + 1}$.
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Limit is 0; test inconclusive. Highest degree terms show limit is 0, test inconclusive.
Limit is 0; test inconclusive. Highest degree terms show limit is 0, test inconclusive.
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Determine divergence: $\lim_{n \to \infty} (-1)^n$.
Determine divergence: $\lim_{n \to \infty} (-1)^n$.
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Does not exist; series diverges. Oscillating sequence has no limit, so series diverges.
Does not exist; series diverges. Oscillating sequence has no limit, so series diverges.
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Evaluate $\lim_{{n \to \infty}} \frac{1}{n^3}$. Divergent?
Evaluate $\lim_{{n \to \infty}} \frac{1}{n^3}$. Divergent?
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Limit is 0; test inconclusive. Higher power in denominator gives limit 0.
Limit is 0; test inconclusive. Higher power in denominator gives limit 0.
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Assess divergence: $\lim_{n \to \infty} \frac{n+1}{n+2}$.
Assess divergence: $\lim_{n \to \infty} \frac{n+1}{n+2}$.
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Limit is $1$; series diverges. Both numerator and denominator approach $\infty$, limit is $1$.
Limit is $1$; series diverges. Both numerator and denominator approach $\infty$, limit is $1$.
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Evaluate $\lim_{n \to \infty} \frac{n}{n+1}$ for divergence.
Evaluate $\lim_{n \to \infty} \frac{n}{n+1}$ for divergence.
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Limit is $1$; series diverges. Since limit equals $1 \neq 0$, the test proves divergence.
Limit is $1$; series diverges. Since limit equals $1 \neq 0$, the test proves divergence.
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Evaluate divergence: $\lim_{n \to \infty} \frac{5n}{3n}$
Evaluate divergence: $\lim_{n \to \infty} \frac{5n}{3n}$
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Limit is $\frac{5}{3}$; series diverges. Coefficients cancel to give limit $\frac{5}{3} \neq 0$
Limit is $\frac{5}{3}$; series diverges. Coefficients cancel to give limit $\frac{5}{3} \neq 0$
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What is the nth term test for divergence?
What is the nth term test for divergence?
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If $\lim_{n \to \infty} a_n \neq 0$, then $\sum a_n$ diverges. The fundamental condition for proving a series diverges.
If $\lim_{n \to \infty} a_n \neq 0$, then $\sum a_n$ diverges. The fundamental condition for proving a series diverges.
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What does $\lim_{{n \to \infty}} a_n = L \neq 0$ imply?
What does $\lim_{{n \to \infty}} a_n = L \neq 0$ imply?
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Series $\sum a_n$ diverges. Any non-zero limit guarantees series divergence.
Series $\sum a_n$ diverges. Any non-zero limit guarantees series divergence.
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Does $\lim_{{n \to \infty}} a_n = 0$ guarantee convergence?
Does $\lim_{{n \to \infty}} a_n = 0$ guarantee convergence?
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No, it does not guarantee convergence. Zero limit is necessary but not sufficient for convergence.
No, it does not guarantee convergence. Zero limit is necessary but not sufficient for convergence.
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What conclusion if $\lim_{{n \to \infty}} a_n = 0$?
What conclusion if $\lim_{{n \to \infty}} a_n = 0$?
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Test is inconclusive. Zero limit cannot determine convergence or divergence.
Test is inconclusive. Zero limit cannot determine convergence or divergence.
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Does $\lim_{{n \to \infty}} a_n = d \neq 0$ infer divergence?
Does $\lim_{{n \to \infty}} a_n = d \neq 0$ infer divergence?
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Yes, series $\sum a_n$ diverges. Any non-zero constant limit proves divergence.
Yes, series $\sum a_n$ diverges. Any non-zero constant limit proves divergence.
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Determine $\lim_{{n \to \infty}} \frac{n + 5}{n}$. Divergent?
Determine $\lim_{{n \to \infty}} \frac{n + 5}{n}$. Divergent?
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Limit is 1; series diverges. Adding constant to numerator gives limit 1.
Limit is 1; series diverges. Adding constant to numerator gives limit 1.
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Determine divergence: $\lim_{{n \to \infty}} \sqrt{n}$.
Determine divergence: $\lim_{{n \to \infty}} \sqrt{n}$.
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Limit is $\infty$; series diverges. Square root grows without bound to infinity.
Limit is $\infty$; series diverges. Square root grows without bound to infinity.
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What does $\lim_{{n \to \infty}} a_n = 0$ indicate?
What does $\lim_{{n \to \infty}} a_n = 0$ indicate?
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Test is inconclusive. Zero limit cannot prove convergence or divergence alone.
Test is inconclusive. Zero limit cannot prove convergence or divergence alone.
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Identify $\lim_{{n \to \infty}} \frac{4n^3}{2n^3}$. Divergent?
Identify $\lim_{{n \to \infty}} \frac{4n^3}{2n^3}$. Divergent?
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Limit is 2; series diverges. Coefficients simplify to $\frac{4}{2} = 2 \neq 0$.
Limit is 2; series diverges. Coefficients simplify to $\frac{4}{2} = 2 \neq 0$.
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Evaluate divergence: $a_n = \sin(n)$.
Evaluate divergence: $a_n = \sin(n)$.
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$\lim_{{n \to \infty}} a_n$ does not exist; series diverges. Sine function oscillates, so limit doesn't exist.
$\lim_{{n \to \infty}} a_n$ does not exist; series diverges. Sine function oscillates, so limit doesn't exist.
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Determine divergence: $\lim_{{n \to \infty}} \frac{n^2 + 2n}{n}$.
Determine divergence: $\lim_{{n \to \infty}} \frac{n^2 + 2n}{n}$.
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Limit is $\infty$; series diverges. Simplifies to $n + 2$, which grows to infinity.
Limit is $\infty$; series diverges. Simplifies to $n + 2$, which grows to infinity.
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What must be true for the nth term test to be applicable?
What must be true for the nth term test to be applicable?
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$a_n$ must be the general term of a series. The test applies only to terms of infinite series.
$a_n$ must be the general term of a series. The test applies only to terms of infinite series.
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Evaluate $\lim_{{n \to \infty}} \frac{n^2 + 1}{n^2}$. Divergent?
Evaluate $\lim_{{n \to \infty}} \frac{n^2 + 1}{n^2}$. Divergent?
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Limit is 1; series diverges. The constant term 1 makes limit equal to 1.
Limit is 1; series diverges. The constant term 1 makes limit equal to 1.
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Determine $\lim_{{n \to \infty}} \frac{2n}{n+1}$. Does it diverge?
Determine $\lim_{{n \to \infty}} \frac{2n}{n+1}$. Does it diverge?
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Limit is $2$; series diverges. Leading coefficients give limit $2 \neq 0$, proving divergence.
Limit is $2$; series diverges. Leading coefficients give limit $2 \neq 0$, proving divergence.
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Determine divergence: $\lim_{{n \to \infty}} \frac{n}{n^2}$.
Determine divergence: $\lim_{{n \to \infty}} \frac{n}{n^2}$.
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Limit is 0; test inconclusive. Denominator grows faster, giving limit 0.
Limit is 0; test inconclusive. Denominator grows faster, giving limit 0.
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Evaluate $\lim_{{n \to \infty}} \frac{n^3}{n^2 + 1}$. Divergent?
Evaluate $\lim_{{n \to \infty}} \frac{n^3}{n^2 + 1}$. Divergent?
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Limit is $\infty$; series diverges. Degree of numerator exceeds denominator, so limit is infinite.
Limit is $\infty$; series diverges. Degree of numerator exceeds denominator, so limit is infinite.
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Check divergence: $\lim_{{n \to \infty}} e^{-n}$.
Check divergence: $\lim_{{n \to \infty}} e^{-n}$.
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Limit is 0; test inconclusive. Exponential decay gives limit 0.
Limit is 0; test inconclusive. Exponential decay gives limit 0.
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Find $\lim_{{n \to \infty}} \frac{1}{n}$ for the nth term test.
Find $\lim_{{n \to \infty}} \frac{1}{n}$ for the nth term test.
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Limit is 0; test inconclusive. This gives the harmonic series limit of 0.
Limit is 0; test inconclusive. This gives the harmonic series limit of 0.
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Evaluate $\lim_{{n \to \infty}} \frac{1}{n^2}$ for divergence.
Evaluate $\lim_{{n \to \infty}} \frac{1}{n^2}$ for divergence.
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Limit is 0; test inconclusive. Zero limit means we need other tests to determine behavior.
Limit is 0; test inconclusive. Zero limit means we need other tests to determine behavior.
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What does $\lim_{{n \to \infty}} a_n = 0$ suggest?
What does $\lim_{{n \to \infty}} a_n = 0$ suggest?
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Test is inconclusive. Zero limit means the test provides no information.
Test is inconclusive. Zero limit means the test provides no information.
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