Sequences and series convergence – university Year 1 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. Dividing top and bottom by \(n\), \(a_n=\dfrac{2-1/n}{1+1/n}\to2\), so \(L=2\) (1). Then \(a_n-2=\dfrac{2n-1-2n-2}{n+1}=\dfrac{-3}{n+1}\), so \(|a_n-L|=\dfrac3{n+1}\) (1).
   2. Let \(\varepsilon>0\) be given and choose an integer \(N\ge3/\varepsilon\) (1). For every \(n>N\), \(|a_n-2|=\dfrac3{n+1}\lt \dfrac3N\le\varepsilon\) (1). Such an \(N\) exists for every \(\varepsilon\), so \(a_n\to2\) by definition.

   In (b), \(N\) depends on \(\varepsilon\): for example, \(\varepsilon=0.01\) needs \(N\ge300\), and indeed \(|a_{301}-2|=\tfrac3{302}\lt0.01\). Note that \(\tfrac3{n+1}\lt\tfrac3N\) because \(n+1\gt N\). The sequence is increasing, since \(a_n=2-\tfrac3{n+1}\).

2. Answer to Question 2 (4 marks)

   1. By partial fractions, \(\dfrac4{(2n-1)(2n+1)}=\dfrac2{2n-1}-\dfrac2{2n+1}\) (1). In the partial sum each negative term cancels the next positive one, leaving \(2-\dfrac2{2N+1}\), which tends to \(2\) (1).
   2. This is a geometric series with first term 3 and ratio \(r=-\frac12\); since \(|r|\lt 1\) it converges (1). Its sum is \(\dfrac3{1-(-1/2)}=\dfrac3{1+1/2}=2\) (1).

   In (a), this is a telescoping series; check the first partial sum, \(\tfrac4{1\times3}=\tfrac43=2-\tfrac23\). In (b), the partial sums 3, 1.5, 2.25, 1.875 oscillate towards 2.

3. Answer to Question 3 (4 marks)

   1. Since \(n^3+1>n^3\), we have \(0\lt \dfrac n{n^3+1}\lt \dfrac n{n^3}=\dfrac1{n^2}\) (1). The series \(\sum\frac1{n^2}\) converges (\(p=2>1\)), so the given series converges by the comparison test (1).
   2. Use the ratio test: \(\dfrac{a_{n+1}}{a_n}=\dfrac{2^{n+1}}{(n+1)^2}\cdot\dfrac{n^2}{2^n}=2\left(\dfrac n{n+1}\right)^2\to2>1\) (1), so the series diverges (1). In fact the terms tend to infinity.

   In (a), the limit comparison test with \(\tfrac1{n^2}\) also works, since \(\tfrac{n}{n^3+1}\div\tfrac1{n^2}\to1\). In (b), the divergence test alone is enough, because \(\tfrac{2^n}{n^2}\) does not tend to zero.

4. Answer to Question 4 (4 marks)

   1. The series of absolute values is \(\sum\frac1{n^2}\), a \(p\)-series with \(p=2>1\), which converges. So the series converges absolutely (2), and absolute convergence implies convergence.
   2. The first four terms give \(1-\frac14+\frac19-\frac1{16}\approx0.7986\) (1). The series alternates with terms decreasing to 0, so the error is less than the next term, \(\frac1{25}=0.04\) (1). The true value \(\frac{\pi^2}{12}\approx0.8225\) confirms this.

   The actual error is about \(0.024\), well within the bound of \(0.04\).

5. Answer to Question 5 (4 marks)

   1. Apply the ratio test to the absolute values: \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac n{n+1}\cdot\dfrac{|x-1|}2\to\dfrac{|x-1|}2\) (1). The series converges when this is less than 1, that is \(|x-1|\lt 2\), so \(R=2\) (1), centred at \(x=1\).
   2. The ratio test is inconclusive at the endpoints, so test them directly. At \(x=3\) the series is \(\sum\frac1n\), the harmonic series, which diverges (1). At \(x=-1\) it is \(\sum\frac{(-1)^n}n\), which converges by the alternating series test. The interval of convergence is \([-1,3)\) (1).

Total : 20 marks