Sequences and Series: math test solutions, College – download the PDF

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Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

Sommaire

(a) Divide by \(n^2\): the limit is \(\frac42=2\). (1 pt)

(b) \(\frac{n}{n+1}\to1\), so the even terms tend to \(1\) and the odd terms to \(-1\). The sequence diverges. (1 pt)

(c) Multiply by the conjugate: \(c_n=\dfrac{n}{\sqrt{n^2+n}+n}=\dfrac{1}{\sqrt{1+\frac1n}+1}\) (1 pt), which tends to \(\tfrac12\). (1 pt)

(a) \(r=-\tfrac23\), \(|r| < 1\), sum \(=\dfrac{5}{1+\frac23}=3\). (1 pt)

(b) \(0.45+0.0045+\cdots\): \(\dfrac{0.45}{0.99}=\dfrac{45}{99}=\dfrac5{11}\). (1 pt)

(c) It converges when \(\left|\tfrac x4\right| < 1\), i.e. \(-4 < x < 4\), and the sum is \(\dfrac{1}{1-\frac x4}=\dfrac{4}{4-x}\). (1 pt)

(a) Limit comparison with \(\frac1{n^2}\): \(\dfrac{5n^3}{n^3+2}\to5\), and \(\sum\frac1{n^2}\) converges, so the series converges. (2 pts)

(b) \(\frac{n}{5n-1}\to\frac15\ne0\): the series diverges by the divergence test. (1 pt)

(a) \(\dfrac{a_{n+1}}{a_n}=\dfrac{2}{n+1}\to0 < 1\) (1 pt), so the series converges. (1 pt)

(b) \(\sqrt[n]{a_n}=\dfrac{n}{2n+1}\to\dfrac12 < 1\) (1 pt), so the series converges. (1 pt)

The terms \(\frac1{2n+1}\) are positive, decreasing and tend to 0, so the series converges. (1 pt)

Limit comparison of \(\sum\frac1{2n+1}\) with \(\sum\frac1n\) gives the limit \(\frac12\), so the absolute series diverges. (1 pt)

The series converges conditionally. (1 pt)

(a) Put \(u=-x^2\): \(\dfrac{1}{1+x^2}=\sum_{n=0}^{\infty}(-1)^n x^{2n}=1-x^2+x^4-\cdots\). (1 pt)

(b) It converges when \(|-x^2| < 1\), so \(|x| < 1\): \(R=1\). (1 pt)

Test yourself: quick challenge for College