Absolute and Conditional Convergence

Theory

  • Absolute Convergence: A series an\sum a_n is absolutely convergent if an\sum |a_n| converges.
  • Conditional Convergence: A series an\sum a_n is conditionally convergent if an\sum a_n converges but an\sum |a_n| diverges.

Ratio Test

Let an\sum a_n be an infinite series and suppose

L=limnan+1an.L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|.
  • If L<1, the series converges absolutely.
  • If L>1, the series diverges.
  • If L=1L=1, the test is inconclusive.

Exercises

Exercise 1
Question

Check whether the series converges:

n=1cos(nπ/3)n2\sum_{n=1}^\infty \frac{\cos(n\pi/3)}{n^2}
Show solution ↓
Solution

Since cos(nπ/3)1|\cos(n\pi/3)|\le 1,

cos(nπ/3)n21n2.\left|\frac{\cos(n\pi/3)}{n^2}\right|\le \frac{1}{n^2}.

Because 1n2\sum \frac{1}{n^2} converges, the given series converges absolutely.

Conclusion: Absolutely convergent.

Exercise 2
Question

Check whether the series converges:

n=1(5)nn!\sum_{n=1}^\infty \frac{(-5)^n}{n!}
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Solution

Apply the Ratio Test:

\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| =\lim_{n\to\infty}\frac{5}{n+1}=0<1.

Conclusion: Absolutely convergent.

Exercise 3
Question

Check whether the series converges:

n=1nsin(1n)\sum_{n=1}^\infty n\sin\left(\frac{1}{n}\right)
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Solution

Using the limit limx0sinxx=1\lim_{x\to 0}\frac{\sin x}{x}=1,

limnnsin(1n)=10.\lim_{n\to\infty} n\sin\left(\frac{1}{n}\right)=1\neq 0.

By the Divergence Test, the series diverges.

Conclusion: Divergent.

Exercise 4
Question

Check whether the series converges:

n=1(1)n+1n\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\sqrt{n}}
Show solution ↓
Solution

Let an=1na_n=\frac{1}{\sqrt{n}}. Then ana_n is decreasing and

limnan=0.\lim_{n\to\infty} a_n=0.

By the Alternating Series Test, the series converges.

However,

n=1(1)n+1n=n=11n\sum_{n=1}^\infty \left|\frac{(-1)^{n+1}}{\sqrt{n}}\right| =\sum_{n=1}^\infty \frac{1}{\sqrt{n}}

diverges.

Conclusion: Conditionally convergent.